46 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Maximum Minimum Values of y on either side of stationary point Both values of y are smaller. Both values of y are larger. Sign of dy dx when moving through a stationary point. + 0 – – 0 + Change of slope of tangent Type of curve Point of inflexion The point C which we considered earlier was a rather special kind of point of inflexion. In general, it is not necessary for the curve at a point of inflexion to have a zero gradient. A point of inflexion is defined simply as a point on a curve at which the tangent to the curve crosses the curve. C P Q Figure 2.12 The point C is of course a point of inflexion, but it is not essential for the gradient to be zero at a point of inflexion. Points P and Q are points of inflexion and the gradients at P and Q are in fact positive and negative respectively. Gradients can be positive, negative or zero. Consider stationary points of inflexion at which dy dx = 0. On either side of such points, the gradient has the same sign so that it goes through the sequence + 0 + or – 0 –, as shown in Figure 2.13. + + 0 – – 0 Figure 2.13
47 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 If the gradient of the point of inflexion is not zero, how then, are we going to find where such points of inflexion occur? Consider the graphs of the gradients. 0 + + – + – + – + – – + – 0 0 y x x x A B x1 x2 x1 x2 x1 x2 dy – dx d 2 y – dx 2 Figure 2.14 P and Q are points of inflexion. In curve A, the gradient is always positive, dy dx reaches a minimum value but not zero. In curve B, the gradient is always negative, dy dx reaches a maximum value but not zero. For both points of inflexion, i.e. at x = x1 and x = x2 , d2 y dx2 = 0. Hence, where points of inflexion occur, d2 y dx2 = 0. But it is possible for d2 y dx2 to be zero at points other than points of inflexion. Thus, if we solve d2 y dx2 = 0, we cannot as yet be sure whether the solution x = a gives a point of inflexion or not. Let us consider two points A, a true point of inflexion, and B, a point on y = f(x) as shown below. B is not a point of inflexion.
48 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 0 + + + + + + – – – 0 0 y A B x x x x3 x4 x3 x4 x3 x4 dy – dx d 2 y – dx 2 Figure 2.15 Note the differences between the two second derived curves. d2 y dx2 is zero for each curve but (a) for point of inflexion A, the graph of d2 y dx2 crosses the x-axis, (b) for point B (not a point of inflexion) the graph of d2 y dx2 only touches the x-axis and the sign of d2 y dx2 does not change. Hence, for a point of inflexion, d2 y dx2 = 0 and there is a change of sign for d2 y dx2 as we go through the point. Summary To find where points of inflexion occur, (a) differentiate y = f(x) twice to get d2 y dx2 , (b) solve the equation d2 y dx2 = 0, (c) test to see if there is any change of sign in d2 y dx2 as we go through this value of x. Higher derivatives Consider the function y = f(x). The first derivative is obtained by differentiating y w.r.t. x, i.e. dy dx = f(x). The second derivative is obtained by differentiating dy dx w.r.t. x, i.e. d dx 1 dy dx 2 = f(x) d dx 1 dy dx 2 is written as d2 y dx2 and is read as ‘dee two y dee x squared’
49 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Note: d2 y dx2 ≠ 1 dy dx 2 2 Hence, if y = f(x), dy dx = f(x) d2 y dx2 = d dx [f(x)] = f(x) d3 y dx3 = d dx [f(x)] = f (x) If y = 2x4 – 5x2 dy dx = 8x3 – 10x d2 y dx2 = 24x2 – 10 d3 y dx3 = 48x If we consider the gradient of the first derived curve and plot this against x, we obtain the second derived curve, which shows values of d2 y dx2 against x. 0 0 y 0 x x x x1 x2 x3 x1 x2 x3 x1 x2 x3 dy – dx d 2 y – dx 2 y = f(x) y = ffi(x) y = ffifi(x) A B C Figure 2.16 From the first derived curve, we see that for stationary points, dy dx = 0. From the second derived curve, we see that, for maximum y, d2 y dx2 is negative, for maximum y, d2 y dx2 is positive, for point of inflexion, d2 y dx2 is zero
50 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Using the second derivative We can use the second derivative to identify the nature of a stationary point instead of looking at the sign of dy dx on either side of it. The next example illustrates the use of the second derivative to identify the nature of the stationary points. Example 24 Given that y = x4 + 4x3 – 2 (a) find dy dx and find the values of x for which dy dx = 0, (b) find the values of d2 y dx2 at each stationary point and hence, determine its nature, (c) find the y values of each of the stationary points. Solution: (a) y = x4 + 4x3 – 2 dy dx = 4x3 + 12x2 = 4x2 (x + 3) dy dx = 0 when x = –3 or 0 (b) d2 y dx2 = 12x2 + 24x When x = 0, d2 y dx2 = 0, so we need to investigate the sign of dy dx on either side of the point. Values of x –1 0 +1 Sign of dy dx + 0 + Change of slope of tangent / — / The sign of dy dx does not change as we go through x = 0. Hence x = 0 is a point of inflexion. When x = –3, d2 y dx2 = 12(–3)2 + 24(–3) = 36 0 d2 y dx2 0 ⇒ a maximum (c) When x = 0, y = –2 so (0, –2) is a point of inflexion. When x = –3, y = (–3)4 + 4(–3)3 – 2 = –29 Thus, (–3, –29) is a minimum point.
51 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Note: 1. d2 y dx2 = 0 does not necessarily indicate that the stationary point is a point of inflexion. It can be a maximum or minimum point. The nature of the stationary point has to be determined by investigating the sign of dy dx on either side of the point. 2. On occasions when it is difficult to find d2 y dx2 , remember that you can always classify the nature of a stationary point by looking at the sign of dy dx on either side of the point. Example 25 Find the stationary points on the graph of the function y = x3 – 3x2 + 2. Distinguish between them. Find the point of inflexion, if any. Solution: y = x3 – 3x2 + 2 dy dx = 3x2 – 6x d2 y dx2 = 6x – 6 Stationary points occur at dy dx = 0 3x2 – 6x = 0 3x(x – 2) = 0 x = 0 or x = 2 When x = 0, y = 2 When x = 2, y = 8 – 3(4) + 2 = –2 Stationary points are (0, 2) and (2, –2). At x = 0, d2 y dx2 = 6(0) –6 = –6 0 Hence, (0, 2) is a maximum point. At x = 2, d2 y dx2 = 6(2) – 6 = 6 0 Hence, (2 –2) is a minimum point. For a point of inflexion, d2 y dx2 = 0 Hence, 6x – 6 = 0 x = 1 To test for a change of sign: Take a point just before x = 1, i.e. x = 1 – a and a point just after x = 1, i.e. x = 1 + a, where a is a small positive quantity. At x = 1 – a, d2 y dx2 = 6(1 – a) – 6 = 6 – 6a – 6 = –6a 0 At x = 1 + a, d2 y dx2 = 6(1 + a) – 6 = 6 + 6a – 6 = 6a . 0
52 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 There is a change of sign in d2 y dx2 as we go through x = 1. Thus, there is a point of inflexion at x = 1. When x = 1, y = 0 Point of inflexion is (1, 0). Alternatively, d3 y dx3 = 6 ≠ 0 Thus, the point of inflexion is (1, 0). To find the stationary point on the curve and to determine whether it is a maximum point, minimum point or a point of inflexion, follow these procedures. 1. Solve the equation dy dx = 0 to find the x-coordinate of the stationary point. 2. Find the y-coordinate by substituting the value of x into y = f(x). 3. Find the value of d2 y dx2 at the stationary point. If d2 y dx2 0, then the point is a minimum point. If d2 y dx2 0, then the point is a maximum point. If d2 y dx2 = 0, determine the nature of the point by investigating the sign of dy dx on either side of the point. Exercise 2.12 1. Find the coordinates of the stationary points of each of the following curves and determine their nature. (a) y = x3 – x2 (b) y = x3 – 6x2 + 12x + 2 (c) y = x(6 – x) 2 (d) y = 3x4 + 4x3 2. Find the turning point of the curve y = 2x + 27 x2 and show that it is a minimum point. 3. Find all the points of inflexion for each of the following curves. (a) y = x3 (b) y = x3 – 6x2 + 5 (c) y = x4 – 4x3 – 18x2 + 7x – 5 (d) y = x4 – 6x3 + 12x2 – 8x 4. Find the coordinates of the point on the curve y = ln x x , (x 0) where dy dx = 0. Determine whether it is a maximum point, a minimum point or a point of inflexion. 5. Find the stationary points on the curve y = xe–2x2 and determine their nature. 6. (a) Find the turning points on the curve y = x3 + 2 x . (b) Show that the gradient of the curve is negative for all negative values of x. 7. The derived function of f(x) = ax2 + bx + c is 4x + 2. Given that the function has a minimum value 1, find the values of a, b and c.
53 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 8. Calculate the coordinates of the maximum and minimum points on the curve y = x + 1 – x2 , distinguishing between maximum and minimum points. 9. Show that the curve x3 + y3 = 3xy has a turning point at (2—1 3 , 2—2 3 ). 10. Differentiate the function x + 3 1 + x2 with respect to x. Find the value of x at which the function has a stationary value and determine the nature of the stationary value. 11. If y = e2x (ax + b) where a and b are constants, find dy dx and d2 y dx2 . Find the values of constants p and q such that d2 y dx2 + p dy dx + qy = 0 for all a and b. 12. If y = x – 1n (1 + ex ), show that (1 + ex ) d2 y dx2 + ex dy dx = 0. 13. If y = (ax + b) e–2x , show that d2 y dx2 + 4 dy dx + 4y = 0. Curve sketching If, for y = f(x), the function f(x) is known, the graph of the function can be plotted by finding the x-coordinates and y-coordinates of a number of selected points. However, this can be quite tedious. Considerable information about the shape and positioning of the curve can be obtained by a systematic analysis of the given equations. The following is a list of steps we can take. 1 Symmetry Inspect the equation for symmetry: (a) If only even powers of y occur, the curve is symmetrical about the x-axis. (b) If only even powers of x occur, the curve is symmetrical about the y-axis. y x 0 y 2 = 4x y x 0 y =x 2 + 1 1 Figure 2.17 Figure 2.18 (c) If only even powers of y and also only even powers of x occur, then the curve is symmetrical about both axes. For example, x2 + y2 = 4 is symmetrical about both axes. y x –2 0 –2 2 2 Figure 2.19
54 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 2 Intersection with the axes Points at which the curve crosses the x and y axes. Crosses the x-axis: Put y = 0 and solve for x Crosses the y-axis: Put x = 0 and solve for y Example: For the curve y(y – 2) = x – 1 When y = 0, x = 1 When x = 0, y(y – 2) = –1 y2 – 2y + 1 = 0 (y – 1)2 = 0 y = 1 Hence, the curve crosses the x-axis at (1, 0) and the y-axis at (0, 1). 3 Stationary points Find the stationary points, i.e. when dy dx = 0 and determine their nature. dy dx = 0 and d2 y dx2 0, then the point is a maximum point. dy dx = 0 and d2 y dx2 0, then the point is a minimum point. dy dx = 0 and d2 y dx2 = 0 with a change of sign through the stationary point, then the point is a point of inflexion. 4 Asymptotes An asymptote to a curve is a straight line to which the curve approaches as the distance from the origin increases. It can also be thought of as a tangent to the curve at infinity. (a) For y = 2 – 1 x , when x → ±∞, the curve approaches the straight line y = 2. Hence, the line y = 2 is an asymptote to the curve. (b) For y = 1 x – 3 , when y → ±∞, the curve approaches the straight line x = 3. Therefore the line x = 3 is an asymptote to the curve. 2 0 y = 2 y x 0 3 x = 3 y x Figure 2.20 Figure 2.21
55 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Asymptote parallel to the axes (a) Rewrite the equation so that it does not contain any algebraic fractions. For example, y = x – 3 x – 2 becomes y(x – 2) = (x – 3). (b) Equate to zero the coefficient of the highest power of y to find the asymptote parallel to the y-axis. (c) Equate to zero the coefficient of the highest power of x to find the asymptote parallel to the x-axis. As an example, find the asymptotes parallel to the axes for the curve y = (x – 2)(x + 3) (x + 1)(x – 4) Rewrite the equation as y(x + 1)(x – 4) = (x – 2)(x + 3) Asymptote parallel to the y-axis: (x + 1)(x – 4) = 0 x = –1 and x = 4 Rearranging the equation gives (y – 1)x2 – (3y + 1)x – 4y + 6 = 0 Asymptote parallel to the x-axis: y – 1 = 0 y = 1 5 Regions of existence of a function Restrictions on the possible range of values that x or y may have. For example, consider y2 = x – 3 Since y2 > 0, ∴ x – 3 > 0 x > 3. Hence, y is defined only for x > 3. 6 Investigate the behaviour of the curve for large values of x and y If the curve does not have any asymptote, then the behavior of y as x → ±∞ is important. For example, (a) if y = x2 – 4 as x → ±∞, y → ∞ (b) if y = x x + 1 = 1 1 + 1 x as x → ±∞, 1 x → 0, y → 1 (c) if y = xex2 , ex2 is always positive. As x → ±∞, y → ±∞. –1 4 1 y x (x – 2)(x + 3) (x + 1)(x – 4) y = – Figure 2.22
56 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Example 26 Determine the turning points and point of inflexion for the curve y = 2x3 + 3x2 – 12x + 7. Sketch the curve. Solution: y = 2x3 + 3x2 – 12x + 7 dy dx = 6x2 + 6x – 12 d2 y dx2 = 12x + 6 At the turning points, dy dx = 0. i.e. 6x2 + 6x – 12 = 0 x2 + x – 2 = 0 (x – 1)(x + 2) = 0 x – 1 = 0 or x + 2 = 0 x = 1 or x = –2 When x = 1, y = 0 d2 y dx2 = 12(1) + 6 = 18 ( 0) Hence, (1, 0) is a minimum point. When x = –2, y = 2(–2)3 + 3(–2)2 – 12(–2) + 7 = 27 d2 y dx2 = –24 + 6 = –18 ( 0) Hence, (–2, 27) is a maximum point. At the point of inflexion, d2 y dx2 = 0. i.e. 12x + 6 = 0 x = – 1 2 When x = – 1 2 , y = 21– 1 2 2 3 + 31– 1 2 2 2 – 121– 1 2 2 + 7 = 13 1 2 . Hence, 1– 1 2 , 13 1 2 2 is a point of inflexion. The sketch of y = 2x3 + 3x2 – 12x + 7 is shown below. x y –2 –1 0 1 27 (–2, 27) – 1_ 2 (1, 0) y = 2x 3 + 3x 2 – 12x + 7 (– , ) _1 2 27_ 2
57 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Example 27 Sketch the curve whose equation is y = x – 3 x – 2 . Solution: Let us work in detail and consider the list of steps. (a) Symmetry: rewrite the equation: y(x – 2) = x – 3 Only odd powers of x and y occur. Thus, no symmetry about the axes. (b) Intersection with the axes. When x = 0, y = 3 2 When y = 0, x = 3 The curve crosses the x-axis at (3, 0) and the y-axis at 10, 3 2 2. (c) Stationary points. dy dx = (x – 2)(1) – (x – 3)(1) (x – 2)2 = x – 2 – x + 3 (x – 2)2 = 1 (x – 2)2 There are no real values of x for dy dx = 0. Therefore, there are no stationary points on the graph. (d) Asymptotes. (i) Asymptote parallel to the y-axis: x – 2 = 0 x = 2 (ii) Rearranging the equation gives (y – 1)x – 2y + 3 = 0. Asymptote parallel to the x-axis: y – 1 = 0 y = 1 The sketch of the curve y = x – 3 x – 2 is shown below. x y 0 1 2 3 x = 2 y = 1 x – 3 x – 2 3_ y = 2 –
58 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Example 28 For the curve y = xe–x , determine the value of x when (a) dy dx = 0, (b) d2 y dx2 = 0. Sketch the curve, showing the turning point, point of inflexion and the general shape of the curve for large values of x and y. Solution: y = xe–x dy dx = e–x + x(–e–x) = e–x (1 – x) d2 y dx2 = –e–x (1 – x) + (–1)(e–x ) = –e–x (1 – x + 1) = –e–x (2 – x) = e–x (x – 2) (a) When dy dx = 0, e–x (1 – x) = 0 i.e. (1 – x) = 0, (e–x ≠ 0) x = 1 Hence, x = 1 when dy dx = 0. (b) When d2 y dx2 = 0, e–x (x – 2) = 0 i.e. (x – 2) = 0, (e–x ≠ 0) x = 2 Hence, x = 2 when d2 y dx2 = 0. When x = 1, y = 1 e d2 y dx2 = e–1 (1 – 2) = – 1 e (, 0) Hence, the curve has a maximum point at 11, 1 e 2. When x = 2, y = 2e–2 = 2 e2 d2 y dx2 = 0 Hence, the curve has a point of inflexion at 12, 2 e2 2. As x → ∞, y = xe–x → 0. As x → –∞, y = xe–x → –∞. The sketch of the curve y = xe–x is shown below. x y 0 y = xe–x _2 e (2, ) _1 e (1, ) 1 2
59 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Example 29 Sketch the curve with the equation y = 1 x ln x. Show clearly on the curve the turning points. Solution: ln x is not defined for x 0, hence the curve y = 1 x ln x does not exist for x 0. y = 1 x ln x dy dx = – 1 x2 ln x + 1 x2 = 1 x2 (1 – ln x) d2 y dx2 = – 2 x3 (1 – ln x) – 1 x3 When dy dx = 0, 1 x2 (1 – ln x) = 0 i.e. (1 – ln x) = 0, 1 1 x2 ≠ 02 ln x = 1 x = e When x = e, y = 1 e ln e = 1 e Hence, the stationary point is 1e, 1 e 2. When x = e, d2 y dx2 = – 2 e3 (1 – ln e) – 1 e3 = 0 – 1 e3 = – 1 e3 ( 0) Hence, 1e, 1 e 2 is a maximum turning point. When d2 y dx2 = 0, – 2 x3 (1 – ln x) – 1 x3 = 0 i.e. – 2 x3 (1 – ln x) = 1 x3 ln x = 3 2 x = e —3 2 When x = e—3 2 , y = 3 2e—3 2 Hence, the point of inflexion is 1 e —3 2 , 3 2e—3 2 2 . As x → ∞, y → 0. As x → 0, 1 x → ∞ but ln x → –∞, ∴ y → –∞.
60 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 When x = 1, ln x = 0, ∴ y = 0. The sketch of the curve y = 1 x ln x is shown below. ( , ) x y 0 1 2 3 e (e, 1 ) _ 1 _ e e y = 1 In x _ x – 3 2e 3 2 e - e 3 2 - 3 2 - Example 30 For the curve y = x2 + 8 1 – x , x ≠ 1, (a) state the asymptote of the equation, (b) show that there is only one turning point and determine its nature, (c) find the coordinates of the point of inflexion. Hence, sketch the curve. Solution: y = x2 + 8 1 – x (a) The asymptote is x = 1. (b) dy dx = 2x + 8 (1 – x) 2 = 2x(1 – x) 2 + 8 (1 – x) 2 = 2x(1 – 2x + x2 ) + 8 (1 – x) 2 = 2x – 4x2 + 2x3 + 8 (1 – x) 2 = 2(x3 – 2x2 + x + 4) (1 – x) 2 = 2(x + 1)(x2 – 3x + 4) (1 – x) 2 At the turning point, dy dx = 0. 2(x + 1)(x2 – 3x + 4) = 0 2(x + 1)3x2 – 3x + 1– 3 2 2 2 – 1– 3 2 2 2 + 44 = 0 2(x + 1)31x – 3 2 2 2 – 9 4 + 44 = 0 2(x + 1)31x – 3 2 2 2 + 7 4 4 = 0 x + 1 = 0 and 1x – 3 2 2 2 + 7 4 ≠ 0 x = –1 Hence, 1x – 3 2 2 2 + 7 4 0
61 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 When x = –1, y = (–1)2 + 8 1 – (–1) = 1 + 8 2 = 5 Hence, there is only one turning point, i.e. (–1, 5). d2 y dx2 = 2 + 16 (1 – x) 3 When x = –1, d2 y dx2 = 2 + 16 [1 – (–1)]3 = 2 + 16 8 = 4 ( 0) Hence, (–1, 5) is a minimum turning point. (c) At the point of inflexion, d2 y dx2 = 0 2 + 16 (1 – x) 3 = 0 2(1 – x) 3 + 16 = 0 2[(1 – x) 3 + 8] = 0 (1 – x) 3 + 8 = 0 (1 – x) 3 = –8 1 – x = 3 –8 1 – x = –2 x = 3 When x = 3, y = 32 + 8 1 – 3 = 9 + (–4) = 5 Hence, the point of inflexion is (3, 5). When x = 0, y = 02 + 8 1 – 0 = 8 As x → ±∞, y = ∞ The sketch of the curve y = x2 + 8 1 – x is shown below. x y 0 –1 1 2 8 3 4 8 (–1, 5) (3, 5) x =1 1 – x y = x 2 + –
62 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Reciprocal graph Consider a curve with equation y = f(x). The reciprocal graph of the curve is given by y = 1 f(x) . We can deduce the graph of y = 1 f(x) from the graph of y = f(x) by following these steps. 1 For each given value of x, f(x) and 1 f(x) have the same sign. 2 If the numerical value of f(x) increases, the numerical value of 1 f(x) will decrease and vice versa. 3 If f(x) → ±∞, 1 f(x) → 0 and if f(x) → 0, 1 f(x) → ±∞. 4 If f(x) has a maximum turning point, then 1 f(x) has a minimum turning point and vice versa. 5 If f(x) = 1, then 1 f(x) = 1 and if f(x) = –1, then 1 f(x) = –1. Example 31 Sketch the graph of the function f(x) = 3 + 2x – x2 . Hence, deduce the graph of the function 1 f(x) . Solution: f(x) = 3 + 2x – x2 = (3 – x)(1 + x) The sketch of the function f(x) = 3 + 2x – x2 is shown below. y x f(x) = 3 + 2x – x 2 1 0 –1 1 3 2 3 4 1 f(x) = 1 3 + 2x – x2 = 1 (3 – x)(1 + x)
63 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 To deduce the graph of 1 f(x) , the following points are noted. (a) As x → –1 or 3, f(x) → 0. Hence, 1 f(x) → ±∞. Thus, x = –1 and x = 3 are asymptotes of the graph of 1 f(x) . (b) As x → ±∞, f(x) → –∞, hence, 1 f(x) → 0. (c) When x = 1, f(x) has a maximum value so 1 f(x) has a minimum value. (d) For x –1, f(x) is negative and the gradient of f(x) is positive, so 1 f(x) is negative and the gradient of 1 f(x) is negative. (e) For –1 x 3, f(x) is positive and the gradient of f(x) is positive, so 1 f(x) is positive and the gradient of 1 f(x) is negative. (f) For 1 x 3, f(x) is positive and the gradient is negative, so 1 f(x) is positive and the gradient is positive. (g) For x 3, f(x) is negative and the gradient is negative, so 1 f(x) is negative and the gradient is positive. Hence, the graph of 1 f(x) = 1 3 + 2x – x2 is shown below. y x 0 1 – 3 1 –1 2 3 x = –1 x = 3 Exercise 2.13 1. Write down the equations of the asymptotes of each of the following curves. (a) y = 2 1 – x (b) y = x 1 – x (c) y = x + 1 x2 (d) y = x – 3 x – 4 (e) y = x 2x + 3 2. Determine the behaviour of y as x → ∞ and x → –∞ if
64 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 (a) y = x 1 + x2 (b) y = x2 e–x (c) y = x ln x (d) y = x3 + x – 3 (e) y2 = 4ax 3. Sketch the graphs of the following curves, showing clearly any asymptotes, turning points, points of intersection with the axes and the behaviour of the curve when x and/or y are very large. (a) y = 2 1 – x (b) y = x 1 – x (c) y = x – 3 x – 4 (d) y = x2 ex (e) y = x3 + 2x2 – x – 2 (f) y = 3x x – 4 (g) y = ex x 4. Find the turning points on the curve with the equation y = 4e4x + 9e–x . Hence, sketch the curve. 5. Find the x-coordinate of the point on the curve y = ln x x2 (x 0) for which dy dx = 0 and determine whether it is a maximum or minimum point. Sketch the curve for x 0. [You may assume that y → 0 as x → ∞] 6. For the curve with the equation y2 = x(2 – x) 2 , (a) state the axis of symmetry of the curve, (b) show that there is no point on the curve for which x is negative, (c) find the coordinates of the points on the curve at which the tangents are parallel to the x-axis, (d) sketch the curve. 7. Sketch the graph of the following curves on separate diagrams. (a) y = x2 + 2 (b) y = 1 x2 + 2 State the coordinates of the turning points for both curves. For (b), show the behaviour of the curve for large positive values of x and large negative values of y. 8. (a) Find the equations of the asymptotes of the curve whose equation is y = x + 2 x – 3 . (b) Find the points of intersection of the curve with the coordinate axes. Find also the stationary points of the curve. (c) Sketch the curve. 9. If y = 3(x – 2) x(x + 6) , find dy dx and deduce the values of x when dy dx = 0. Determine the nature of these points. Sketch the graph showing the above properties and the asymptotes. 10. Find the coordinates of the turning point on the curve y = ex + 2e–x and show that it is a minimum turning point. Sketch the curve. 11. Given that f(x) = 6x2 + x – 12, find the minimum value of f(x) and the values of x for which f(x) = 0. Using the same axes, sketch the curves y = f(x) and y = 1 f(x) , labelling each curve clearly.
65 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 12. Given that y = x – 3 x – 4 , (a) find dy dx , (b) find the equation of the tangent to the curve at the point (6, 1.5), (c) find the equation of the normal to the curve at the point (5, 2), (d) use your answer from (a) to deduce that the curve has no turning points and sketch the graph. 13. For the curve y = xp e –—1 2 x , where p is an integer greater than 1, find the value of p when (a) dy dx = 0, (b) d2 y dx2 = 0. Sketch the graphs of y = x2 e –—1 2 x and y = x3 e –—1 2 x , showing clearly the turning point, point of inflexion and the behaviour of the curve when x → ±∞. Equation of the tangent and the normal to a curve Equation of the tangent to the curve The gradient of a curve at a point is the gradient of the tangent to the curve at that point. If y = f(x) represents the equation of a curve, then dy dx is the gradient of the tangent to the curve at that point. If the coordinates of the point are known, we can find the equation of the tangent. Example 32 Find the equation of the tangent to the curve y = (x – 2)(x2 + 1) at the point where x = –1. Solution: y = (x – 2)(x2 + 1) = x3 – 2x2 + x – 2 dy dx = 3x2 – 4x + 1 When x = –1, dy dx = 8 When x = –1, y = (–3)(2) = –6 ∴ The gradient of the tangent at (–1, –6) is 8. Equation of the tangent to the curve at the point (–1, –6) is given by y – y1 = m(x – x1 ) y – (–6) = 8[x – (–1)] y = 8x + 2
66 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Example 33 A curve has parametric equations x = a cos3 q and y = sin3 q where a is a positive constant and 0 < q < 2π. Find the equation of the tangent at the point (a cos3 q, a sin3 q), where q is a parameter. Solution: Given y = a sin3 q and x = a cos3 q ∴ dy dq = 3a sin2 q cos q and dx dq = –3a cos2 q sin q dq dx = – 1 3a cos2 q sin q dy dx = dy dq × dq dx = – 3a sin2 q cos q 3a cos2 q sin q = – sin q cos q = –tan q Hence the equation of the tangent at the point (a cos3 q, a sin3 q) is y – a sin3 q = –tan q (x – a cos3 q) y – a sin3 q = – sin q cos q (x – a cos3 q) y cos q – a cos q sin3 q = –x sin q + a sin q cos3 q Multiplying both sides by cos q y cos q + x sin q = a cos q sin3 q + a sin q cos3 q y cos q + x sin q = a cos q sin q (sin2 q + cos2 q) y cos q + x sin q = a sin q cos q Equation of the normal to the curve The normal to a curve y = f(x) at a particular point is the straight line which is at right angles to the tangent at the point. (Figure 2.23) If the gradient of the tangent is m1 , the gradient, m2 , of the normal is given by m2 = – 1 m1 . This enables us to find the equation of the normal at any specific point on a curve. Example 34 Find the equation of the normal to the curve y = x2 + 5x – 2 at the point where the curve meets the line x = 4. Solution: y = x2 + 5x – 2 dy dx = 2x + 5 When x = 4, dy dx = 2(4) + 5 = 13 When x = 4, y = 42 + 5(4) – 2 = 34 0 Tangent Normal A y y = f(x) x Figure 2.23
67 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Gradient of the tangent at the point (4, 34) is 13 and gradient of the normal at the point (4, 34) is – 1 13 . Equation of the normal to the curve at the point (4, 34) is given by y – y1 = m(x – x1 ) y – 34 = – 1 13 (x – 4) 13y – 442 = –x + 4 13y + x – 446 = 0 Example 35 Find the equation of the normal to the curve 2x2 – xy + 3y2 = 5 at the point (2, 1). Solution: 2x2 – xy + 3y2 = 5 Differentiate w.r.t. x, 4x – (x dy dx + y) + 6y dy dx = 0 (6y – x) dy dx = y – 4x dy dx = y – 4x 6y – x When x = 2, y = 1, dy dx = 1 – 8 6 – 2 = – 7 4 The gradient of the tangent at the point (2, 1) is – 7 4 and the gradient of the normal at the point (2, 1) is 4 7 . Equation of the normal to the curve at the point (2, 1) is given by y – y1 = m(x – x1 ) y – 1 = 4 7 (x – 2) 7y – 7 = 4x – 8 7y = 4x – 1 Exercise 2.14 1. Find the equations of the tangent and the normal to each of the following curves at the given point. (a) y = x2 – 2 at the point (1, –1) (b) y = x2 – 5x + 2 at the point (3, –4) (c) y = (2x – 1)(x + 1) at the point (1, 1) (d) y = (x – 2)(x2 + 1) at the point (–1, –6) (e) y = 1 x at the point (–1, –1)
68 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 2. Find the equation of the tangent to the curve y = (x – 5)(2x + 1) which is parallel to the x-axis. 3. Find the value of k such that the line y = 2x + k is a normal to the curve y = 2x2 – 3. 4. Find the equation of the normal to the curve y = x2 + 3x – 2 at the point where the line crosses the y-axis. 5. Find the equations of the tangents to the curve y = (2x – 1)(x + 1) at the point where the curve crosses the x-axis. Find the point of intersection of the two tangents. 6. A curve has equation y = x2 – 5x + 6. Find the equations of the normal to the curve at the point where the curve crosses the x-axis. 7. Find the coordinates of the point on the curve at which the gradient is 3. Hence, find the value of c such that the line y = 3x + c is a tangent to the curve y = x2 – 5. 8. A curve has equation y = x2 – 3x + 2. Find the equation of the normal to the curve which has a gradient of 1 2 . 9. A curve has parametric equations x = t 2 , y = 1 – 1 t (t 0). The curve crosses the x-axis at A. Find the equation of the tangent to the curve at the point A. 10. The parametric equations of a curve are x = 2t 2 and y = t 3 . Express dy dx in terms of t. Find the equation of the tangent to the curve at the point Q(2q2 , q3 ). 11. A curve has parametric equations x = t(t 2 + 1) and y = t 2 + 1. Find in its simplest form, the equation of the tangent to the curve at the point with parameter t. 12. Find the equation of the normal to the curve x = cos q, y = sin q at the point where q = π 4 . Find the coordinates of the point on the curve where the normal meets the curve again. Rate of change Rates of change can be expressed using differentials. If y = f(x), dy dx measures the rate of change of y with respect to x. Rates of increase are positive. Rates of decrease are negative. Rates are often, but not always, rates of change with respect to time. However, by convention, ‘the rate of change of a quantity Q’ refers to the rate of change of Q with respect to time i.e. dQ dt . Rates of change can be related by the chain rule dy dt = dy dx · dx dt It enables us to find the rate of change of y with respect to t, if y is a function of x and the rate of change of x with respect to t is known. This is useful when solving problems concerning rates of change of physical quantities.
69 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Example 36 A spherical balloon is being blown up so that its volume increases at a constant rate of 30 cm3 s–1. When the radius is 12 cm, at what rate is the radius increasing? Solution: Let the volume of the balloon be V and its radius be r at time t, V = 4 3 πr3 Differentiate w.r.t. r, dV dr = 4πr2 Using the chain rule, dV dt = dV dr × dr dt = 4πr2 × dr dt dr dt = 1 4πr 2 × dV dt When r = 12 and dV dt = 50, dr dt = 1 4π(12)2 · 50 = 0.028 Thus, the rate of increase of the radius is 0.028 cm s–1. Example 37 An inverted right circular cone has a semi-vertical angle of 30°. Water runs out of a small hole at the vertex of the cone at a constant rate of 3 cm3 s–1. Find the rate at which the depth h of the water is falling when the volume of water is 81π cm3 . Solution: Volume of cone, V = 1 3 πr 2 h But r = h tan 30° = h 3 Hence V = 1 3 π 1 h 3 2 2 h = πh3 9 Differentiate w.r.t. h, dV dh = 1 3 πh2 Using the chain rule, dV dt = dV dh × dh dt = 1 3 πh2 · dh dt When V = 81π cm3 , πh3 9 = 81π ⇒ h = 9 Also dV dt = –3, hence –3 = 1 3 π92 · dh dt dh dt = – 1 9π Hence, the rate of decrease of the height is 1 9π cm s–1. h r 30°
70 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Exercise 2.15 1. The side of a cube is increasing at a constant rate of 0.1 m s–1. Find the rate of increase of its volume when each side is 2 m long. 2. The volume of a spherical balloon is increasing at a constant rate of 0.05 m3 s–1. Find the rate of increase of its radius when the volume of the balloon is 0.008 m3 . 3. A spherical balloon is being inflated in such a way that its volume is increasing at a constant rate of 8 cm3 s–1. Find the rate of increase of the surface area of the balloon when its radius is 10 cm. 4. The radius of a circular oil slick is increasing at 1.5 m s–1. Taking π to be 3.14, find, to 2 significant figures, the rate at which the area of the oil slick is increasing when its radius is 300 m. 5. An inverted right circular cone of semi-vertical angle 45° is collecting water from a tap at a steady rate of 18π cm3 s–1. Find the rate at which the depth h of the water is rising when h = 3 cm. 6. A hemispherical bowl of radius 6 cm contains water which is flowing into it at a constant rate. When the height of the water is h cm, the volume V of the water in the bowl is given by V = π16h2 – 1 3 h3 2 cm3 . Find the rate at which the water level is rising when h = 3, given that the time taken to fill the bowl is 1 minute. 7. A ladder 5 m long is leaning against a vertical wall. The bottom of the ladder is pulled along the ground away from the wall at a constant rate of 0.4 m s–1. How fast will the top of the ladder be falling at the instant when its bottom is 3 m away from the wall? 8. A flask in the shape of a cone of height 20 cm and radius 8 cm is held vertex downwards. Show that when the depth of water in the flask is x cm, the volume of water is 4 75 πx3 cm3 . Water leaks out from the vertex at the rate of 2 cm3 s–1. Find the rate of change of the depth of the water when the depth is 10 cm. 9. A hemispherical bowl of radius 12 cm is initially full of water. Water runs out of a small hole at the bottom of the bowl at a rate of 48π cm3 s–1. When the depth of the water is x cm, show that the depth is decreasing at a rate of 48 x(24 – x) cm s–1. Find the rate at which the depth is decreasing when (a) the bowl is full, (b) the depth is 6 cm. [When the depth of water in the hemispherical bowl with radius r is x cm, the volume of the water is 1 3 πr2 (3r – x) cm3 ] 10. Two variables p and q are connected by the relation 1 p + 1 q = 1 k where k is a constant. Given that k = 10 and p decreases at the rate of 2 cm s–1, find the rate of change of q when p = 40.
71 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Optimisation problems The methods used to find maximum and minimum points can be applied to practical problems in which the maximum and minimum values of a quantity are required. The procedures are as follow. 1. Write an expression for the required quantity. 2. Use the given conditions to rewrite it in terms of a single variable. 3. Find the turning points and their nature. It is often obvious from the problem itself whether a maximum or minimum has been obtained. Example 38 A rectangle has perimeter 28 m. What is its maximum area? Solution: Let x m and y m be the length of the sides of the rectangle. Its perimeter = 2x + 2y = 28 x + y = 14 y = 14 – x Area of rectangle, A = xy = x(14 – x) Rewirte so that A has only one variable. Differentiate A w.r.t. x, dA dx = 14 – 2x. When A is maximum, dA dx = 0 i.e. 14 –2x = 0 x = 7 d2 A dx2 = –2 0 ⇒ maximum Since y = 14 – x = 7, maximum area, A = 7 × 7 m2 = 49 m2 . Example 39 A right circular cone of height (a + x), where –a x a, is inscribed in a sphere of fixed radius a, so that the vertex and all points of the circumference of the base are on the surface of the sphere. Show that the volume of the cone is given by V = 1 3 π (a – x)(a + x) 2 . Find the maximum volume of V as x varies. Solution: Base radius of cone, r = a2 – x2 using Phythagoras theorem Volume of cone, V = 1 3 πr2 h, h = height of cone = 1 3 π(a2 – x2 )(a + x) = 1 3 π(a – x)(a + x) 2 dV dx = – 1 3 π(a + x) 2 + 1 3 π(a – x)(a + x) = 1 3 π(a + x)(a – 3x)
72 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 For maximum or minimum V, dV dx = 0 1 3 π(a + x)(a – 3x) = 0 x = 1 3 a, (x ≠ –a) d2 V dx2 = 1 3 π(a – 3x) – π(a + x) = 1 3 π(–2a – 6x) = – 2 3 π(a + 3x) When x = 1 3 a, d2 V dx2 = – 4 3 πa 0 Hence, the volume is maximum when x = 1 3 a Vmax = π 3 (a – a 3 )(a + a 3 ) 2 = 32 81 πa3 . Exercise 2.16 1. Find the minimum value of x2 + 2y2 if x and y are related by x + 2y = 1. 2. If xy = 4, find the maximum and minimum values of P = x + y where x and y are positive. 3. If V = x2 h and h + x = 6, find the maximum and minimum values of V. 4. A window consists of a rectangle of width 2x and height a, surrounded by a semicircle of diameter 2x. If the perimeter of the window is P, show that the area of the window is given by A = Px – 2x2 – 1 2 πx2 . If x varies and P is a constant, find, in terms of P, the greatest possible area of the window. 5. A solid circular cylinder has a given volume. Show that its total surface area will be least when its height is equal to the diameter of the base. 6. The lengths of the sides of a rectangular sheet of metal are 8 cm and 3 cm. A square of side x cm is cut from each corner of the sheet and the remaining piece is folded to make an open box. (a) Show that the volume V of the box is given by V = 4x3 – 22x2 + 24x. (b) Find the value of x for which the volume of the box is a maximum. Calculate the maximum volume. 7. An open rectangular box is made of very thin sheet metal. Its volume is 128 cm3 , its width is x cm and its length of 4x cm. Obtain an expression for its depth in terms of x. Show that the total surface area of its base, its ends and its sides is equal to (4x2 + 320 x ) cm2 . Calculate the dimensions of the box for which the surface area is a minimum.
73 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 8. The diagram on the right shows the junction of two roads which are perpendicular to each other. The width of the roads are 9 m and 4 m respectively. S and T are two variable points such that SBT is a straight line. If CT = x m and AS = y m, express y in terms of x and find the value of x such that TA + AS is a minimum. 9. The diagram on the right shows the cross-section of a cylindrical solid with radius r and height h. A hemisphere with radius r is attached to the upper surface of the cylinder. If the volume of the combined solid is V, show that the total surface area of the combined solid is 2V r + 5πr 2 3 . If r varies and V is a constant, prove that the surface area is minimum when r = h. 10. A rectangular garden consists of a rectangular lawn of area 72 m2 surrounded by a concrete path. The path is 2 m wide at two opposite edges of the garden and 1 m wide along each of the other two edges. Find the dimensions of the garden of smallest area satisfying these requirements. Summary 1. The derivative of a function f with respect to x is defined by f(x) = lim δx → 0 f(x + x) – f(x) δx where x is a small change in the value of x. 2. Derivative of standard functions. Function Derivative constant 0 xn nxn – 1 ex ex ax ax ln a ln x 1 x sin x cos x cos x –sin x tan x sec2 x sin–1 x 1 (1 – x2 ) cos–1 x – 1 (1 – x2 ) tan–1 x 1 1 + x 2 4 cm 9 cm S A T x y C B h j
74 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 3. (a) Differentiation of sums and differences y = u ± v where u and v are functions of x. dy dx = du dx ± dv dx (b) Differentiation of products y = uv where u and v are functions of x dy dx = u dv dx + v du dx (c) Differentiation of quotients y = u v where u and v are functions of x dy dx = v du dx – u dv dx v2 (d) Differentiation of composite functions If y is a function of u and u is a function of x, then dy dx = dy du × du dx 4. Parametric differentiation If y = f(t) and x = g(t), then dy dx = dy dt × dt dx . 5. dy dx is the gradient of a curve. 6. If y = f(x), the rate of change of y is given by dy dt = dy dx × dx dt . 7. For maximum point, (a) values of y on either side of the stationary point are smaller, (b) dy dx changes from positive to negative when moving through the stationary point, (c) d2 y dx2 0. 8. For minimum point, (a) values of y on either side of the stationary point are larger, (b) dy dx changes from negative to positive when moving through the stationary point, (c) d2 y dx2 0. 9. For point of inflexion, (a) dy dx has the same sign on either side of the stationary point, i.e. + 0 + or – 0 –. (b) d2 y dx2 = 0 and there is a change of sign of d2 y dx2 (i.e. d3 y dx3 ≠ 0). 10. Equations of tangent and normal (a) The equation of the tangent to the curve at the point P(x1 , y1 ) is y – y1 = m(x – x1 ), where m = dy dx , i.e. the gradient of the tangent at point P. (b) The equation of the normal to the curve at the point P(x1 , y1 ) is y – y1 = m1 (x – x1 ), where m1 , is the gradient of the normal at P and mm1 = –1. 11. A curve y = f(x) is (a) concave upwords if f'(x) increases as x increases or f"(x) . 0, (b) concave downwards if f'(x) decreases as x increases or f"(x) , 0.
75 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 STPM PRACTICE 2 1. Differentiate each of the following functions with respect to x and simplify your answers. (a) x4 – 2x + 1 x2 (b) x2 – 2x – 4 x2 – 4 (c) 2x – 3 4 x (d) (x + 1)—3 2 (x – 2)—1 2 (e) ln sin2 3x (f) etan x (g) (x – 2)3 e–5x (h) ln1 2 – x2 1 + x 2 (i) sin x cos3 x (j) e–2x (3 sin 3x – 2 cos 3x) (k) 2x – 3 4x2 + 9 (l) 2x – 1 x + 2 (m) tan2 x sin 3x (n) x4 + 1 1 + x (o) e2x 1 + 2e3x 2. Given that y = x cos x + sin x x2 , find dy dx and simplify your answer. 3. Given that y = ex – e–x ex + e–x , show that dy dx = 1 – y2 . 4. Find the value of dy dx at the point (2, 4) on the curve xy2 = 16. 5. If x2 + xy + y2 = 7, find the value of dy dx when x = 3 and y = –1. 6. If f(x) = ln (1 + x) – x + 1 2 x2 , find f(x). If x 0, then f(x) 0. 7. For the curve with equation x3 + 2y3 + 3xy = 0, find the gradient at the point (2, –1). 8. Given that y3 e–2x2 = xe4x , prove that dy dx = y(2x + 1)2 3x . 9. Given that y = cos ln (1 + x), prove that (a) (1 + x) dy dx = –sin ln (1+ x) (b) (1 + x2 ) d2 y dx2 + (1 + x) dy dx + y = 0 10. Given that y = sin x x2 , x 0, prove that x2 d2 y dx2 + 4 dy dx + (x2 + 2)y = 0. 11. If x = 2t t + 2 and y = 3t t + 3 , find the value of dy dx at the point 1 2 3 , 3 4 2.
76 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 12. The parametric equations of a curve are x = q – sin q and y = 1 – cos q, where 0 q 2π. Show that dy dx = cot q 2 and find the equation of the tangent and the normal to the curve at q = π 2 . 13. A curve has parametric equations x = 2t – ln 2t and y = t 2 – 1n t 2 , where t 0. Find the value of t at the point on the curve at which the gradient is 2. 14. If y3 = 6xy – x3 – 1, prove that dy dx = 2y – x2 y2 – 2x and that the maximum value of y occurs when x3 = 8 + 2 114 and the minimum value when x3 = 8 – 2 114 . 15. The equation of a curve is y = 1 x – 32x2 . (a) Find the coordinates of the stationary point and determine its nature. (b) Determine the coordinates of the inflexion point. (c) Sketch the curve. 16. Find the equation of the normal to the curve x3 – 4x2 y + 3xy2 + y3 = 27 at the point where the curve intersects with y-axis. 17. The parametric equations of a curve are x = 2t + 3t 2 and y = 1 – 2t 2 . Find an equation of the tangent to the curve which is parallel to the line 2y + x = 0. 18. The parametric equations of a curve are x = t 2 + t and y = t 2 – t. Find dy dx in terms of y. The curve crosses the y-axis at A where t 0. Find the equation of the normal to the curve at A. 19. (a) The curve y = x2 + ax + b has a turning point at (1, 3). Find a and b. (b) Find the value of k if x (x + 1)2 (x – k) has a stationary point. 20. Find the maximum point on the curve y = 1 x2 + 2x + 4 and show that there are points of inflexion at 10, 1 4 2 and 1–2, 1 4 2. 21. (a) Find the stationary point of the curve y = 4x2 + 1 x and show that this is a minimum point. (b) The curve y = x(1 + x)(3 – x) has a stationary point. Find the value of x. Show that the tangent to the curve at the point (1, 4) meets the curve again at the origin. 22. Show that the curve y = 3x4 – 8x3 – 6x2 + 24x + k, where k is a constant, has a stationary point at x = 1. Hence, find the values of x of another two stationary points on the curve. 23. A spherical balloon is inflated by gas being such that its volume is increasing at a constant rate. Show that the rate of increase of the surface area of the balloon is inversely proportional to its radius. (Volume of sphere = 4 3 πr3 , surface area of sphere = 4πr2 .) 24. An empty container is being filled with a liquid. The height of the liquid in the container at time t seconds is x cm and the volume of the liquid is V cm3 where V = π 2 x(x + 4). Given that V increases at a constant rate and x = 8 when t = 16, find (a) the rate of change of the volume in terms of π, (b) the rate of increase of x when x = 10.
77 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 25. A spherical raindrop is formed by condensation. In an interval of 40 seconds, its volume increases at a constant rate from 0.032 mm3 to 0.256 mm3 . Find the rate at which the surface area of the raindrop is increasing when its radius is 0.5 mm. 26. A container full of milk takes the form of an inverted right circular cone of height 10 cm and base radius 4 cm. Find, in cm s–1, the rate at which the milk level in the container is falling when the height of milk in the container is 5 cm, given that milk is flowing from the container at the rate of 100 cm3 s–1. 27. A right circular cone of height h is inscribed in a sphere of radius R. Show that the volume V of the cone is given by V = π 2 (2Rh2 – h3 ). 28. A right circular cone of base radius r and height h has a total surface area S and volume V. Show that 9V2 = r 2 (S2 – 2πr 2 S). Hence, or otherwise, show that for a fixed surface area S, the maximum volume of the cone occurs when its semi-vertical angle q is given by tan q = 1 2 2 . (Total surface area of cone = πrl + πr2 where l is the length of the slanting edge of the cone). 29. The diagram shows a house at A, a school at D and a straight road BC, where ABCD is a rectangle with AB = 2 km and BC = 6 km. θ θ 2 km 6 km B P Q C A D A boy travels from A to D by walking to a point P on the road, running along the road to a point Q and then walking from Q to D. The points P and Q are chosen so that the angles APB and DQC are both equal to q. Given that the boy walks at a constant speed of 4 kmh–1 and runs at a constant speed of 8 kmh–1, show that the time, T minutes, taken by the boy to go from A to D is given by T = 15(3 + 4 sin q – 2 tan q ). Show that as q varies, the minimum time for the journey is approximately 97 minutes. 30. A right pyramid has a square base of side x m and a total surface area (base and four sides) 72m2 . Show that the volume, V cm3 , is given by V2 = 144x2 – 4x4 . If x varies, find the value of x for which V is a maximum and obtain the maximum value of V. [Volume of pyramid = 1 3 (base area × height)] 31. If y = e–x 1 + x2 , prove that (a) y is always positive, (b) dy dx is never positive. Find the coordinates of the point where dy dx = 0 and sketch the graph of the curve. (It can be assumed that as x → –∞, y → +∞)
78 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 32. (a) For the curve y = (x + 1)2 (1 – x), find the x-coordinate of the turning point and the point of inflexion. Sketch the curve for –2 x 2. (b) Sketch the curve y2 = (x + 1)2 (1 – x) for –2 x 2 on separate diagram, paying particular attention to its behaviour when y = 0. 33. If y2 = x2 (x – 2), obtain an expression for dy dx in terms of x. Deduce that the curve has no turning points. Show that when x = 2 2 3 , dy dx = ± 6 and d2 y dx2 = 0. 34. Water is poured at a constant rate of 20 cm3 s –1 into an empty inverted conical funnel of radius 20 cm and height 50 cm. (a) Determine the rate of change of the height of water in the funnel, given that h is the height of water in centimeter at time t seconds. (b) Calculate the rate of change of the radius of water in the funnel when the radius of the surface of water is half of the radius of the funnel. (c) Find the time taken to fill the inverted funnel completely. 35. (a) Find the coordinates of the turning points for the curve y = x2 (2 – x) and y = 1 x2 (2 – x) . (b) Sketch the two curves on separate diagrams and state the equations of the asymptotes of the second curve. (c) Find the set of values of k such that the equation x2 (2 – x) = k has real roots. 36. Sketch the curve with equation y = x 2 – x and state (a) the equation of each of the asymptotes, (b) an equation of the tangent at the origin. Sketch on a separate diagram, the curve with equation y = x 2 – x . 37. Sketch the curve y = (2x – 1)2 (x + 1), showing the coordinates of (a) the points where it meets the axes, (b) the turning points, (c) the point of inflexion. Hence, sketch the graph of y = 1 (2x – 1)2 (x + 1) , showing the coordinates of any turning points clearly. 38. Given x . 0 and f(x) = 1 x , find lim h → 0 f(x + h) – f(x) h . 39. A curve is defined by the parametric equations x = t – 1 t and y = 3t + 1 t , where t ≠ 0. (a) Show that dy dx = 3 – 4 t 2 + 1 and hence, deduce that –1 , dy dx , 3. (b) Find the coordinates of the points where dy dx = 1.
79 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 40. Given u = 1 2 (ex – e–x ), where x . 0 and y = f(u) is a differentiable function f. If dy du = 1 u2 + 1 , show that dy dx = 1. 41. Given that yx2 = a sin mx, using implicit differentiation, show that x2 d2 y dx2 + 4x dy dx + (m2 x2 + 2)y = 0. 42. The function f is defined by f(x) = ln x x (a) State all asymptotes of f. (b) Find the stationary point of f, and determine its nature. (c) Obtain the intervals, where (i) f is concave upwards, and (ii) f is concave downwards. Hence, determine the coordinates of the point of inflexion (d) Sketch the graph y = f(x). 43. The parametric equations of a curve are x = 1 – cos t, y = t + sin t. Find the equation of the normal to the curve at a point with parameter 1 2 π. 44. A curve is defined implicitly by the equation x2 + xy + y2 = 27. (a) Find dy dx in terms of x and y. (b) Find the gradients of the curve at the points where the curve crosses the x-axis and y-axis. (c) Show that the coordinates of the stationary points of the curve are (–3, 6) and (3, –6) and determine their nature. 45. Find the gradients of the curve xy2 + x2 – 3x2 y = 19 at the points where the x-coordinates is 1. 46. Equation of a curve is f(x) = x3 e –x . (a) Find the coordinates of the stationary points on the curve and determine its nature. (b) Sketch the curve of f(x). 47. A rectangle with a width 4x is inscribed in a circle of constant radius r. (a) Express the area, A of the rectangle in terms of x and r. (b) Determine x in terms of r when area of the rectangle has a maximum value.
CHAPTER INTEGRATION 3 Subtopic Learning Outcome 3.1 Indefinite integrals (a) Identify integration as the reverse of differentiation. (b) Integrate xn (n Q), ex , sin x, cos x, sec2 x, with constant multiples, sums and differences. (c) Integrate rational functions by means of decomposition into partial fractions. (d) Use trigonometric identities to facilitate the integration of trigonometric functions. (e) Use algebraic and trigonometric substitutions to find integrals. (f) Perform integration by parts. 3.2 Define integrals (a) Identify a definite integral as the area under a curve. (b) Use the properties of definite integrals. (c) Evaluate definite integrals. (d) Calculate the area of a region bounded by a curve (including a parametric curve) and lines parallel to the coordinates axes or between two curves. (e) Calculate volumes of solids of revolution about one of the coordinates axes. anti-derivative – kamiran tak tentu area under a curve – luas di bawah lengkung constant of integration – pemalar pengamiran definite integral – kamiran tentu indefinite integral – kamiran tak tentu integral – kamiran integration – pengamiran integration by parts – pengamiran bahagian demi bahagian partial fraction – pecahan separa substitution – gantian volume of revolution – isi padu kisaran Bilingual Keywords
81 Mathematics Semester 2 STPM Chapter 3 Integration 3 3.1 Indefinite Integrals Integration as the Reverse of Differentiation Integration is the reverse process of differentiation. In differentiation, we start with a function and then proceed to find its derivative. In integration, we start with the derivative and then work back to find the function from which it is derived. Given that d dx F(x) = f(x), the process of finding F(x) which has the derivative f(x), is called integration. We write ∫f(x) dx = F(x) + c The expression ∫f(x) dx denotes the indefinite integral of f(x) with respect to x. The expression f(x) to be integrated is called the integrand. F(x) is called an anti-derivative of f(x), c is known as the constant of integration. Notice that d dx (x3 ) = 3x2 ⇒ ∫ 3x2 dx = x3 d dx (x3 + 4) = 3x2 ⇒ ∫ 3x2 dx = x3 + 4 d dx (x3 – 2) = 3x2 ⇒ ∫ 3x2 dx = x3 – 2 In these examples, we happen to know the expressions from which the derivative 3x2 was derived. Any constant term in the original expression becomes zero in the derivative. We therefore acknowledge the presence of such a constant term of some value by adding a symbol c to the result of integration. i.e. ∫ 3x2 dx = x3 + c Example 1 Show that d dx 1 x 1 + 5x 2 = 1 (1 + 5x) 2 Hence, find ∫ 1 (1 + 5x) 2 dx. Solution: d dx 1 x 1 + 5x 2 = 1 + 5x – 5x (1 + 5x) 2 = 1 (1 + 5x) 2 ∫ 1 (1 + 5x) 2 dx = x 1 + 5x + c Integration INFO
82 Mathematics Semester 2 STPM Chapter 3 Integration 3 Example 2 Differentiate e2x (2x – 1) 4 with respect to x. Hence, find ∫ xe2x dx. Solution: Let y = e2x (2x – 1) 4 dy dx = 1 4 · 2e2x (2x – 1) + 1 4 · 2e2x = 1 2 e2x (2x – 1 + 1) = xe2x Hence ∫ xe2x dx = 1 4 e2x (2x – 1) + c Exercise 3.1 1. Differentiate ln (x + x2 + 1 ) with respect to x. Hence, find ∫ 1 x2 + 1 dx. 2. Given that y = (x + 3)2x – 3 , show that dy dx = 3x 2x – 3 . Hence, find ∫ x 2x – 3 dx. 3. Find dy dx (x cos x). Hence, find ∫(cos x – x sin x) dx. 4. Differentiate x2 2x – 1 with respect to x, and hence find ∫ x(x – 1) (2x – 1)2 dx. 5. Given that y = ln cos x, find dy dx . Hence, find ∫tan x dx. Standard integrals Integral of xn Recall that: (a) d dx (xn ) = nxn – 1 (b) d dx (xn + 1) = (n + 1)xn (c) d dx 1 1 n + 1 xn + 12 = xn Reversing this, ∫ xn dx = xn + 1 n + 1 + c, n ≠ – 1. In general, ∫ xn dx = xn + 1 n + 1 + c. Feynman's Integration VIDEO
83 Mathematics Semester 2 STPM Chapter 3 Integration 3 i.e. to integrate a power of x, add 1 to the power and divide by the new power. Note: 1. This is true except when n = –1. 2. When n = 0, ∫ x0 dx = ∫ 1 dx = x + c Example 3 Find the integrals (a) ∫ x6 dx. (b) ∫ 1 x dx Solution: (a) ∫ x6 dx = x6 + 1 6 + 1 + c (b) ∫ 1 x dx = ∫ x 1 –—2 dx = x7 7 + c = x 1 –—2 + 1 – 1 2 + 1 + c = 2x —1 2 + c = 2x + c Integral a constant d dx (ax) = a where a is a constant. Thus ∫ a dx = ax + c Integral of axn d dx (ax n + 1) = a(n + 1)xn Hence ∫ a(n + 1)xn dx = ax n + 1 + c or ∫ axn dx = axn + 1 n + 1 + c Integral of 1 x (a) x 0 Since 1 x ≡ x–1, it may appear that the rule ∫ xn dx = xn + 1 n + 1 + c can be used. But when n = –1, this method fails. Recall that d dx (ln x) = 1 x and that ln x is defined only when x 0. Hence ∫ 1 x dx = ln x + c, if x 0.
84 Mathematics Semester 2 STPM Chapter 3 Integration 3 (b) x 0 If x is negative, ∫ 1 x dx = ln x + c is not valid since the logarithm of a negative number does not exist. But if we write ∫ 1 x dx = ∫ –1 –x dx, (–x 0) = ln (–x) + c Thus, when x 0, ∫ 1 x dx = ln x + c when x 0, ∫ 1 x dx = ln (–x) + c Combining these results, we have ∫ 1 x dx = ln |x| + c, for all values of x. Integral of ex d dx (ex ) = ex ∫ ex dx = ex + c Integrals of sin x, cos x and sec2 x d dx sin x = cos x d dx cos x = –sin x d dx tan x = sec2 x Hence ∫ cos x dx = sin x + c ∫sin x dx = –cos x + c ∫sec2 x dx = tan x + c Basic properties of integration: (a) If a is a constant, ∫ a f(x) dx = a ∫f(x) dx (b) ∫[f(x) ± g(x) dx = ∫f(x) dx ± ∫ g(x) dx
85 Mathematics Semester 2 STPM Chapter 3 Integration 3 Example 4 Find (a) ∫ 5ex dx (b) ∫(6x3 – 2x 2 ) dx (c) ∫(x2 + 5)2 dx (d) ∫(2x – 1)(3x + 2) dx Solution: (a) ∫ 5ex dx = 5 ∫ ex dx = 5ex + c (b) ∫(6x3 – 2x 2 ) dx = ∫ 6x3 dx – ∫ 2x2 dx = 6 4 x4 – 2 3 x3 + c = 3 2 x4 – 2 3 x3 + c (c) ∫(x2 + 5)2 dx = ∫(x4 + 10x2 + 25) dx = ∫ x4 dx + ∫ 10x2 dx + ∫ 25 dx = x 5 5 + 10 3 x3 + 25x + c (d) ∫(2x – 1)(3x + 2) dx = ∫(6x2 + x – 2) dx = ∫ 6x2 dx + ∫ x dx – ∫ 2 dx = 6x 3 3 + x 2 2 – 2x + c = 2x3 + 1 2 x2 – 2x + c Example 5 Find (a) ∫sin2 x dx (b) ∫ cos2 2x dx (c) ∫sin2 x cos2 x dx Solution: (a) cos 2x = 1 – 2 sin2 x sin2 x = 1 2 (1 – cos 2x) ∫sin2 x dx = 1 2 ∫(1 – cos 2x) dx = 1 2 1x – 1 2 sin 2x2 + c = 1 2 x – 1 4 sin 2x + c
86 Mathematics Semester 2 STPM Chapter 3 Integration 3 (b) cos 2x = 2 cos2 x – 1 cos 4x = 2 cos2 2x – 1 cos2 2x = 1 2 (1 + cos 4x) dx ∫ cos2 2x dx = 1 2 ∫(1 + cos 4x) dx = 1 2 1x + 1 4 sin 4x2 + c = 1 2 x + 1 8 sin 4x + c (c) sin 2x = 2 sin x cos x sin x cos x = 1 2 sin 2x ∫sin2 x cos2 x dx = ∫1 sin 2x 2 2 2 dx = 1 4 ∫sin2 2x dx = 1 4 ∫1 1 – cos 4x 2 2 dx = 1 8 ∫(1 – cos 4x) dx = 1 8 1x – 1 4 sin 4x2 + c = 1 8 x – 1 32 sin 4x + c Integrals of the form {[f(x)]n f(x)} If f(x) is a function of x and d dx f(x) = f(x), d dx [f(x)]n + 1 n + 1 = [f(x)]n f(x) Reversing the process, ∫[f(x)]n f(x) dx = [f(x)]n + 1 n + 1 Example 6 Integrate with respect to x. (a) (x – 2)3 (b) 2x(x2 – 3)4 Solution: (a) Let f(x) = x – 2 f(x) = 1 Hence (x – 2)3 is of the form [f(x)]n f(x), where n = 3, f(x) = x – 2 and f(x) = 1. Hence ∫(x – 2)3 dx = (x – 2)4 4 + c.
87 Mathematics Semester 2 STPM Chapter 3 Integration 3 (b) Let f(x) = x2 – 3 f(x) = 2x Thus 2x(x2 – 3)4 is of the form [f(x)]n f(x), where n = 4, f(x) = x2 – 3 and f(x) = 2x. Hence, ∫ 2x(x2 – 3)4 dx = (x2 – 3)5 5 + c Example 7 Find the indefinite integral. (a) ∫ x2 x3 – 5 dx (b) ∫ cos x sin3 x dx (c) ∫ x x2 + 1 dx Solution: (a) x2 x3 – 5 = 1 3 · 3x2 (x3 – 5)—1 2 , which is of the form [f(x)]n f(x), where n = 1 2 , f(x) = x3 – 5, f(x) = 3x2 . Hence, ∫ x2 x3 – 5 dx = 1 3 ∫ 3x2 (x3 – 5)—1 2 dx = 1 3 · (x3 – 5)—3 2 2 3 + c = 2 9 (x3 – 5)—3 2 + c (b) cos x sin3 x is in the form [f(x)]n f'(x), where n = 3, f(x) = sin x and f'(x) = cos x. Hence, ∫ cos x sin3 x dx = sin4 x 4 + c (c) x x2 + 1 = x(x2 + 1)–—1 2 = 1 2 · 2x(x2 + 1)–—1 2 Hence, ∫ x x2 + 1 dx = ∫ 1 2 · 2x(x2 + 1)–—1 2 dx = 1 2 · (x2 + 1)—1 2 1 2 + c = x2 + 1 + c
88 Mathematics Semester 2 STPM Chapter 3 Integration 3 Exercise 3.2 1. Find the indefinite integrals (a) ∫ x5 dx (b) ∫ 3ex dx (c) ∫ 5 sin x dx (d) ∫ x —1 2 dx (e) ∫ x–3 dx (f) ∫ 2 cos x dx (g) ∫ 6 x dx (h) ∫ 1 x 4 dx 2. Find the indefinite integrals (a) ∫(4x2 – 5x3 ) dx (b) ∫ x 2 (x – 2) dx (c) ∫ 5x + 4 x3 dx (d) ∫ x + 1 x dx (e) ∫(4x – 3)2 dx (f) ∫( 3 x – x ) 2 dx (g) ∫(3 + ex )(2 + e–x ) dx (h) ∫(2x – 1 x 2 ) 2 dx 3. Using the formula ∫[f(x)]n f(x) dx = [f(x)]n + 1 n + 1 + c, find (a) ∫(x + 4)5 dx (b) ∫ 3x 2 (x3 – 1)2 dx (c) ∫ xx2 – 1 dx (d) ∫ x2 x3 + 1 dx (e) ∫ In x x dx (f) ∫ ex 1 + ex dx (g) ∫ cos4 x sin x dx (h) ∫tan3 x sec2 x dx (i) ∫ cos x sin4 x dx (j) ∫sin x cos x dx Integration by partial fractions Consider ∫ x + 4 2x2 + x – 1 dx. Clearly this is not a standard integral and the numerator is not the derivative of the denominator. Hence, it is not possible to use any of the integration techniques that you have learnt. So how do we go about solving it? In Chapter 2, you have learnt how to express a rational function in its partial fractions. Thus x + 4 2x2 + x – 1 can, in fact, be expressed as 3 2x – 1 – 1 x + 1 .
89 Mathematics Semester 2 STPM Chapter 3 Integration 3 Therefore ∫ x + 4 2x2 + x – 1 dx = ∫1 3 2x – 1 – 1 x + 1 2 dx = ∫ 3 2x – 1 dx – ∫ 1 x + 1 dx = 3 2 ln |2x – 1| – ln |x + 1| + c This method is known as integration by partial fractions. In general, if the integrand is of the form f(x) g(x) , where f(x) and g(x) are polynomials of x and g(x) can be factorised, f(x) g(x) must first be expressed as partial fractions before integration is attempted. Note: Only proper fractions can be converted directly into partial fractions. An improper fraction must first be divided first to obtain its proper fractions. Example 8 Find ∫ 4x + 3 2x + 1 dx. Solution: ∫ 4x + 3 2x + 1 dx = ∫12 + 1 2x + 1 2 dx (divide out the integrand) = 2x + 1 2 ln |2x + 1| + c Example 9 Find ∫ x + 4 2x2 + x – 1 dx. Solution: First, change the expression into partial fractions. x + 4 2x2 + x – 1 ≡ x + 4 (2x – 1)(x + 1) ≡ A 2x – 1 + B x + 1 x + 4 ≡ A(x + 1) + B(2x – 1) Let x = –1, 3 = –3B ⇒ B = –1 Let x = 1 2 , 9 2 = 3 2 A ⇒ A = 3 Thus, x + 4 2x2 + x – 1 ≡ 3 2x – 1 – 1 x + 1 Hence,∫ x + 4 2x2 + x – 1 dx = ∫1 3 2x – 1 – 1 x + 1 2 dx = ∫ 3 2x – 1 dx – ∫ 1 x + 1 dx = 3 2 ln |2x – 1| – ln |x + 1| + c
90 Mathematics Semester 2 STPM Chapter 3 Integration 3 Example 10 Integrate x2 + 1 x2 – 1 with respect to x. Solution: x2 + 1 x2 – 1 ≡ 1 + 2 x2 – 1 ≡ 1 + 2 (x – 1)(x + 1) Let 2 (x – 1)(x + 1) ≡ A x – 1 + B x + 1 2 ≡ A(x + 1) + B(x – 1) Let x = 1, 2 = 2A ⇒ A = 1 Let x = –1, 2 = –2B ⇒ B = –1 Thus, ∫ x2 + 1 x2 – 1 dx = ∫ 11 + 1 x – 1 – 1 x + 1 2 dx = x + ln |x – 1| – ln |x + 1| + c = x + ln x – 1 x + 1 + c Exercise 3.3 Express the integrand in each of the following integrals in partial fractions, and hence find the integrals. 1. ∫ x x + 2 dx 2. ∫ 1 (1 – x)(3x – 2) dx 3. ∫ 2x + 3 (x – 4)(5x + 2) dx 4. ∫ 1 x2 + 4x + 3 dx 5. ∫ 7x – 2 (x – 2)2 (2x + 3) dx 6. ∫ 1 x2 (1 – x) dx 7. ∫ 2x – 4 (x2 + 4) (x + 2) dx 8. ∫ 4x2 x(2x – 1)2 dx 9. ∫ x2 x + 1 dx 10. ∫ x2 + x + 5 x(x + 1) dx 11. ∫ x2 + 2x + 4 (2x – 1)(x2 – 1) dx 12. ∫ 12x (2 – x)(3 – x)(4 – x) dx 13. Express 1 x2 (x – 1) in the form A x + B x2 + C x – 1 where A, B and C are constants. Hence, find ∫ 1 x2 (x – 1) dx. 14. Using the substitution u = ex , show that ∫ ex – 1 ex + 1 dx = ∫ u – 1 u(u + 1) du. Hence, using partial fractions, find ∫ ex – 1 ex + 1 dx.
91 Mathematics Semester 2 STPM Chapter 3 Integration 3 Integration by substitution We are often required to integrate functions which are in the standard form where x is replaced by a linear function of x. E.g. ∫(5x – 4)6 dx, which resembles ∫ x6 dx except that it is replaced by (5x – 4). Let us substitute u = 5x – 4 and make u the variable of integration. The integral becomes ∫ u6 dx and before we can complete the operation, we must change the variable, thus ∫(5x – 4)6 dx = ∫ u6 dx = ∫ u6 dx du du Now dx du can be found from the substitution u = 5x – 4, for du dx = 5. Therefore, dx du = 1 5 and the integral becomes ∫ u6 dx = ∫ u6 dx du du = ∫ u6 1 1 5 2 du = 1 5 ∫ u6 du = 1 5 1 u7 7 2 + c Finally, we must express u in terms of the original variable x, so ∫(5x – 4)6 dx = (5x – 4)7 35 + c The above method is known as integration by substitution. In general, ∫f(x) dx = ∫ g(u) dx du du Example 11 Using the substitution u = 2x – 1, find the anti-derivative of x(2x – 1)7 . Solution: Let u = 2x – 1 ⇒ x = 1 2 (u + 1) du dx = 2 ⇒ dx du = 1 2 Hence ∫ x(2x – 1)7 dx = ∫ 1 2 (u + 1)(u7 ) 1 2 du = 1 4 ∫(u8 + u7 ) du
92 Mathematics Semester 2 STPM Chapter 3 Integration 3 = 1 4 3 u9 9 + u8 8 4 + c = u8 288 (8u + 9) + c = (2x – 1)8 288 [8(2x – 1) + 9] + c = 1 288 (2x – 1)8 (16x + 1) + c Example 12 Using the substitution u = 3x + 1 dx, find ∫ x 3x + 1 dx. Solution: Let u = 3x + 1 ⇒ u2 = 3x + 1 2u du dx = 3 dx du = 2u 3 ∫ x 3x + 1 dx = ∫ u2 – 1 3u · 2u 3 du = 2 9 ∫(u2 – 1) du = 2 9 ( u3 3 – u) + c = 2 27 u(u2 – 3) + c = 2 27 3x + 1 (3x + 1 – 3) + c = 2 27 3x + 1 (3x – 2) + c Example 13 Find ∫ 1 (9 – 4x2 ) dx using the substitution x = 3 2 sin q. Solution: Let x = 3 2 sin q dx dq = 3 2 cos q
93 Mathematics Semester 2 STPM Chapter 3 Integration 3 ∫ 1 (9 – 4x2 ) dx = ∫ 1 9 – 9 sin2 q · 3 2 cos q dq = ∫ 1 3 cos q · 3 2 cos q dq = ∫ 1 2 dq = 1 2 q + c Hence, ∫ 1 (9 – 4x2 ) dx = 1 2 sin–1 2x 3 + c. Some common types of substitution Integral Substitution 1. ∫(ax + b) n dx u = ax + b 2. ∫ ax + b dx u2 = ax + b 3. ∫ a2 – x2 dx x = a sin q 4. ∫ a2 + x2 dx x = a tan q 5. ∫ 1 a cos x + b sin x dx or ∫ 1 a + b cos x dx or ∫ 1 a + b sin x dx t = tan x 2 6. sinm x cosn x, m, n ∈ Z+ (a) when m is odd (b) when n is odd u = cos x u = sin x Example 14 Using a suitable substitution, find (a) ∫ x (2x + 3)3 dx (b) ∫ cos3 x dx (c) ∫ 4 – x2 dx Solution: (a) Let u = 2x + 3 ⇒ x = u – 3 2 du dx = 2 ⇒ dx du = 1 2
94 Mathematics Semester 2 STPM Chapter 3 Integration 3 ∫ x (2x + 3)3 dx = ∫ u – 3 2u3 . 1 2 du = 1 4 ∫ u – 3 u3 du = 1 4 ∫1 1 u2 – 3 u3 2 du = 1 4 1– 1 u + 3 2u2 2 + c = – 1 4(2x + 3) + 3 8(2x + 3)2 + c (b) Let u = sin x du dx = cos x ⇒ dx du = 1 cos x ∫ cos3 x dx = ∫ cos2 x · cos x dx = ∫ (1 – sin2 x) cos x dx sin2 x + cos2 x = 1 = ∫ (1 – u2 ) cos x · 1 cos x du = ∫ (1 – u2 ) du = u – u3 3 + c = sin x – 1 3 sin3 x + c (c) Let x = 2 sin q dx dq = 2 cos q ∫ 4 – x2 dx = ∫ 4 – 4 sin2 q 2 cos q dq = ∫ 4 1 – sin2 q cos q dq = ∫ 4 cos2 q dq = ∫ 2(1 + cos 2q) dq cos 2q = 2 cos2 q – 1 = 21q + 1 2 sin 2q2 + c = 2q + sin 2q + c = 2q + 2 sin q cos q + c = 2q + 2 sin q 1 – sin2 q + c = 2 sin–1 1 x 2 2 + x 1 – x 2 + c
95 Mathematics Semester 2 STPM Chapter 3 Integration 3 Example 15 Find ∫ 1 1 + sin x dx Solution: Let tan x 2 = t 1 2 sec2 x 2 · dx dt = 1 dx dt = 2 sec2 x 2 = 2 1 + t 2 sec2 x 2 = 1 + tan2 x 2 sin x = 2 sin x 2 cos x 2 cos2 x 2 + sin2 x 2 = 2 tan x 2 1 + tan2 x 2 ÷ cos2 x 2 = 2t 1 + t 2 ∫ 1 1 + sin x dx = ∫ 1 1 + 2t 1 + t 2 · 2 1 + t 2 dt = ∫ 2 t 2 + 2t + 1 dt = 2 ∫ 1 (1 + t) 2 dt = 2 ∫(1 + t) –2 dt = 2(1 + t) –1 –1 + c = – 2 1 + tan x 2 + c Exercise 3.4 1. Find the following integrals using the suggested substitution. (a) ∫ x(x2 + 2)3 dx, u = x2 + 2 (b) ∫ 2x(x2 – 4)—1 2 dx, u = x2 – 4 (c) ∫ x 2x2 – 5 dx, u = 2x2 – 5 (d) ∫ x x + 9 dx, u = x + 9 cos2 x 2 = 1 + sin2 x 2 = 1