96 Mathematics Semester 2 STPM Chapter 3 Integration 3 (e) ∫ 2x + 1 (x – 3)6 dx, u = x – 3 (f) ∫ 3x 4 – x dx, u2 = 4 – x (g) ∫ 1 – x2 dx, x = sin q (h) ∫sin5 x dx, u = cos x (i) ∫sin4 x cos x dx, u = sin x (j) ∫ 1 4 + x2 dx, x = 2 tan q (k) ∫ x2 + 1 x3 + 3x + 4 dx, u = x3 + 3x + 4 (l) ∫ cos3 x sin x dx, u = cos x (m) ∫ 1 + x x dx, u2 = x (n) ∫ 1 (9 – x 2 ) —3 2 dx, x = 3 sin q 2. Using a suitable substitution, find (a) ∫ 2x(1 – x) 7 dx (b) ∫ x 1 – 4x2 dx (c) ∫ x + 3 (4 – x) 5 dx (d) ∫ 1 9 – x2 dx (e) ∫ x2 4 – x2 dx (f) ∫ sin3 x dx (g) ∫sin3 x cos x dx Formulae derived from integration by substitution 1. Integrating (ax + b) n , where a and b are constants. Consider ∫(ax + b) n dx Let u = ax + b du dx = a ⇒ dx du = 1 a ∫(ax + b) n dx = ∫ un · 1 a du = 1 a ∫ un du = un + 1 a(n + 1) + c = (ax + b) n + 1 a(n + 1) + c In general, ∫(ax + b) n dx = (ax + b) n + 1 a(n + 1) + c
97 Mathematics Semester 2 STPM Chapter 3 Integration 3 Example 16 Find (a) ∫ (4x + 7)8 dx (b) ∫ (2 – 1 3 x) 7 dx (c) ∫ 2 – 5x dx Solution: (a) ∫(4x + 7)8 dx = (4x + 7)9 (4)(9) + c = (4x + 7)9 36 + c (b) ∫(2 – 1 3 x) 7 dx = 12 – 1 3 x2 8 1– 1 3 2(8) + c = – 3 8 12 – 1 3 x2 8 + c (c) ∫ 2 – 5x dx = ∫(2 – 5x) —1 2 dx = (2 – 5x) —3 2 (–5)1 3 2 2 + c = – 2 15 (2 – 5x) —3 2 + c 2. Consider ∫(ax + b) n dx. If n = –1, ∫(ax + b) –1 dx = ∫ 1 ax + b dx Let u = ax + b du dx = a Hence, ∫ 1 ax + b dx = ∫ 1 u · 1 a du = 1 a ∫ 1 u du = 1 a ln |u| + c Thus, ∫ 1 ax + b dx = 1 a ln |ax + b| + c
98 Mathematics Semester 2 STPM Chapter 3 Integration 3 Example 17 Find (a) ∫ 1 3x – 2 dx (b) ∫ 2 3 – 4x dx Solution: (a) ∫ 1 3x – 2 dx = 1 3 ln |3x – 2| + c (b) ∫ 2 3 – 4x dx = 2 ∫ 1 3 – 4x dx = 2 –4 ln |3 – 4x| + c = – 1 2 ln |3 – 4x| + c 3. Integrals of the form ∫ f(x) f(x) dx Consider a function lnu where u is a function of x. Differentiating with respect to x, d dx lnu = 1 u du dx = du dx u So ∫ du dx u dx = ln|u| + c If we write f(x) = u and f(x) = du dx , ∫ f(x) f(x) dx = ln |f(x)| + c Example 18 Find (a) ∫ 3x2 x3 – 4 dx (b) ∫ 4x – 8 x2 – 4x + 5 dx (c) ∫ tan x dx (d) ∫ ex ex + 4 dx Solution: (a) Let f(x) = x3 – 4 f(x) = 3x2 ∫ 3x2 x3 – 4 dx is of the form ∫ f(x) f(x) dx Hence, ∫ 3x2 x3 – 4 dx = ln |x3 – 4| + c
99 Mathematics Semester 2 STPM Chapter 3 Integration 3 (b) Let f(x) = x2 – 4x + 5 f(x) = 2x – 4 Hence, ∫ 4x – 8 x2 – 4x + 5 dx = 2 ∫ 2x – 4 x2 – 4x + 5 dx = 2 ln |x2 – 4x + 5| + c (c) tan x = sin x cos x Let f(x) = cos x f(x) = –sin x Hence, ∫ tan x dx = – ∫ –sin x cos x dx = –ln |cos x| + c (d) Let f(x) = ex + c f(x) = ex Hence, ∫ ex ex + 4 dx = ln |ex + 4| + c By using the method of substitution, the following results are obtained. ∫ f(x) ef(x) dx = ef(x) + c ∫ eax + b dx = 1 a eax + b +c ∫sin (ax + b) dx = – 1 a cos (ax + b) + c ∫ cos (ax + b) dx = 1 a sin (ax + b) + c ∫sec2 (ax + b) dx = 1 a tan (ax + b) + c Example 19 Find (a) ∫ e3x – 2 dx (b) ∫sin 5x dx (c) ∫ cos (5 – 3x) dx (d) ∫(ex – 4)2 dx (e) ∫ x2 ex3 dx Solution: (a) ∫ e3x – 2 dx = 1 3 e3x – 2 + c (b) ∫sin 5x dx = – 1 5 cos 5x + c (c) ∫ cos (5 – 3x) dx = – 1 3 sin (5 – 3x) + c
100 Mathematics Semester 2 STPM Chapter 3 Integration 3 (d) ∫(ex – 4)2 dx = ∫(e2x – 8ex + 16) dx = 1 2 e2x – 8ex + 16x + c (e) Let f(x) = x3 f(x) = 3x2 Hence, ∫ x2 ex3 dx = ∫ 1 3 · 3x2 ex3 dx = 1 3 ∫ 3x2 ex3 dx = 1 3 ex3 + c Exercise 3.5 1. Find the following integrals using the formula ∫(ax + b) n dx = (ax + b) n + 1 a(n + 1) + c, n ≠ –1. (a) ∫(3x – 9)9 dx (b) ∫(2 – 5x) 6 dx (c) ∫ 1 (4x + 5)3 dx (d) ∫ 1 4 3 + 5x dx (e) ∫ 1 – x + 1 1 – x – 1 (1 – x) 2 dx 2. Using the formula ∫ eax + b dx = 1 a eax + b + c, find (a) ∫ e2 – 5x dx (b) ∫ ex (e2x + e3x ) dx (c) ∫(e–x + 2)2 dx (d) ∫1ex + 1 ex 2 2 dx (e) ∫ e4x + e–x e3x dx 3. Using the result ∫ f(x) f(x) dx = ln |f(x)| + c, find (a) ∫ 1 2x dx (b) ∫ 1 3x + 2 dx (c) ∫ 1 3 – 2x dx (d) ∫ x + 1 x2 + 2x + 5 dx (e) ∫ x2 x3 + 1 dx (f) ∫ cos x 2 – sin x dx
101 Mathematics Semester 2 STPM Chapter 3 Integration 3 (g) ∫ e3x e3x + 1 dx (h) ∫ 1 x ln x dx (i) ∫ sec2 x 1 + tan x dx (j) ∫ sin x – cos x sin x + cos x dx 4. Using the result ∫f(x) ef(x) dx = ef(x) + c, write down the anti-derivatives (a) sec2 x etan x (b) e —1 x x2 (c) ex x (d) x ex2 5. Write down ∫ ex 1 + ex dx. Show that 1 1 + ex = 1 – ex 1 + ex . Hence, find ∫ 1 1 + ex dx. Integration by parts We often need to integrate a product where either function is not the derivative of the others. For example ∫ x2 ln x dx, ∫ x e–x dx, ∫ ex sin x dx etc. In situations like this, we have to find some other method of dealing with the integral. Consider the product rule for differentiation: d dx (uv) = u dv dx + v du dx where u and v are functions of x. Now integrate both sides with respect to x, uv = ∫ u dv dx dx + ∫ v du dx dx and rearranging the terms, we have ∫ u dv dx dx = uv – ∫ v du dx dx This method is known as integrating by parts. Note: When faced with integrating the product of two functions by means of this formula, we have to split the function into two simpler functions, one called u and the other dv dx . We will need to decide which will be which. (a) As we want to use du dx on the right-hand side of the expression, u should be a function which becomes a simpler function after differentiation. (b) As we need v to work out the right-hand side of the expression, it must be possible to integrate the function dv dx to obtain v.
102 Mathematics Semester 2 STPM Chapter 3 Integration 3 Example 20 Find ∫ 2x ex dx. Solution: First split 2x ex into 2 simpler functions, 2x and ex . Both can be integrated easily but as 2x becomes a simpler function after differentiation and ex does not, take u to be 2x. Let u = 2x ⇒ du dx = 2 and dv dx = ex ⇒ v = ex (we omit the integration constant here as we are in the middle of evaluating the integral) Substituting into ∫ u dv dx dx = uv – ∫ v du dx dx gives ∫ 2x ex dx = 2x ex – ∫ 2ex dx = 2x ex – 2ex + c (The integration constant appears here) Example 21 Find ∫ x2 ln x dx. Solution: We must choose u = lnx since we cannot integrate lnx (it has to be done by parts) Let u = lnx ⇒ du dx = 1 x and dv dx = x2 ⇒ v = x3 3 Substituting into ∫ u dv dx dx = uv – ∫ v du dx dx, ∫ x2 lnx dx = lnx 1 x3 3 2 – ∫ x3 3 · 1 x dx = x3 3 lnx – 1 3 ∫ x2 dx = x3 3 lnx – 1 9 x3 + c = x3 3 1lnx – 1 3 2 + c Note: If one of the factors of the product to be integrated is a log term, this must be chosen as u. Sometimes it is necessary to use integration by parts more than once to complete the integration successfully. Example 22 Find ∫ x2 e3x dx. Solution: First split x2 e3x into x2 and e3x . Since x2 becomes a simpler function after differentiation, take u to be x2 . Let u = x2 ⇒ du dx = 2x and dv dx = e3x ⇒ v = e3x 3
103 Mathematics Semester 2 STPM Chapter 3 Integration 3 Substituting into ∫ u dv dx dx = uv – ∫ v du dx dx, ∫ x2 e3x dx = x2 1 e3x 3 2 – ∫ 1 3 e3x · 2x dx = x2 3 e3x – 2 3 ∫ x e3x dx. The integral ∫ x e3x dx has to be integrated by parts again. Let u = x ⇒ du dx = 1 and dv dx = e3x ⇒ v = 1 3 e3x Thus, ∫ x e3x dx = x 3 e3x – ∫ 1 3 e3x dx = x 3 e3x – 1 9 e3x Hence, ∫ x2 e3x dx = x2 3 e3x – 2 3 ( x 3 e3x – 1 9 e3x ) + c = x2 3 e3x – 2 9 x e3x + 2 27 e3x + c = e3x 3 (x2 – 2 3 x + 2 9 ) + c Example 23 Find ∫ e3x sin x dx. Solution: Let u = e3x ⇒ du dx = 3e3x and dv dx = sin x ⇒ v = –cos x ∫ e3x sin x dx = e3x (–cos x) – ∫(–cos x)(3e3x ) dx = –e3x cos x + 3 ∫ e3x cos x dx and again by parts = –e3x cos x + 3{e3x sin x – 3 ∫sin x(e3x ) dx} and it looks as though we are back to where we started. However, if we write I = ∫ e3x sin x dx I = –e3x cos x + 3e3x sin x – 9I. Then treating this as an equation, we get 10I = e3x (3 sin x – cos x) + c Hence, I = 1 10 e3x (3 sin x – cos x) + c or ∫ e3x sin x dx = 1 10 e3x (3 sin x – cos x) + c
104 Mathematics Semester 2 STPM Chapter 3 Integration 3 Note: Whenever we integrate functions of the form ekx sin x or ekx cos x, we get similar results after applying the rule twice. Summary: From the given examples, we can make the following conclusions. (a) If one factor is a log. function, it must be taken as ‘u’. (b) If there is no log. function but a power of x, that becomes ‘u’. (c) If there is neither a log. function nor a power of x, then the exponential function is taken as ‘u’. Exercise 3.6 1. Write down the function to be taken as u and the function to be taken as dv du ; Use the formula for integration by parts to find the integrats. (a) ∫ x ex dx (b) ∫ x cos 3x dx (c) ∫ x e–2x dx (d) ∫(2x + 1) cos x dx (e) ∫ln x dx (f) ∫ x lnx dx (g) ∫ x sin x dx (h) ∫ x sec2 x dx (i) ∫ln 3x dx 2. Using integration by parts twice in each case, find the anti-derivative of (a) x2 ex (b) x2 e–2x (c) x2 sin x (d) ex sin x (e) x (1nx) 2 3. Find ∫ x 1 + x dx (a) using integration by parts, (b) using the substitution u = 1 + x. 4. Find ∫ 2x(x – 2)4 dx (a) using integration by parts, (b) using the substitution u = x – 2. 3.2 Definite Integrals Definite integral as the area under a curve Consider the area under the curve y = f(x) between x = a and x = b, where f is continuous for a x b. y x 0 a b y = f(x) y x a P Q 0 b y = f(x) δx δy y x x + δx Figure 3.1 Figure 3.2
105 Mathematics Semester 2 STPM Chapter 3 Integration 3 Let P be the point (x, y) on the curve. Q(x + x, y + y) is a point near P on the same curve. The approximate area A of the rectangular strip under the arc PQ is given by A ≈ y x. If we divide the whole area between x = a and x = b into a number of such strips, the total area is approximately the sum of the areas of all the strips. This can be written as A ≈ x = b x Σ = a y x If we make the strips narrower i.e. as dx gets smaller, the accuracy of the results increases. Then, as x → 0 A = lim dx → 0 x = b x Σ = a y x Alternatively, since A ≈ y x A x ≈ y lim dx → 0 A x = y But lim dx → 0 A x = dA dx so dA dx = y Hence A = ∫ y dx. The boundary values of x defining the total area are x = a and x = b. This is written as total area = ∫ a b y dx. Therefore, the total area can be found in two ways 1. by a process of summation, 2. by integration. lim dx → 0 x = b x Σ = a y x = ∫ a b y dx Hence, we can conclude that integration is a process of summation. Definite integrals Let f be a function defined on the interval [a, b]. The definite integral of f from a to b is defined by ∫ a b f(x) dx = lim n → ∞ n i Σ = 1 f(xi ) x, provided the limit exists where x = 1 n (b – a) and xi is any value of x in the ith interval. The numbers a and b are called the limits of integration. Suppose ∫f(x) dx = F(x) + c. Then ∫ a b f(x) dx = 3F(x)4 a b = F(b) – F(a) Note: With a definite integral, the constant of integration may be omitted as it disappears in subsequent working.
106 Mathematics Semester 2 STPM Chapter 3 Integration 3 Example 24 Evaluate (a) ∫ 1 2 (3x – 2) dx (b) ∫ 0 5 1 x + 5 dx (c) ∫ –1 —1 2 4e2x dx Solution: (a) ∫ 1 2 (3x – 2) dx = 3 3 2 x2 – 2x4 1 2 = (6 – 4) – 1 3 2 – 22 = 2 1 2 (b) ∫ 0 5 1 x + 5 dx = 3ln (x + 5)4 0 5 = ln 10 – ln 5 = ln 10 5 = ln 2 (c) ∫ –1 —1 2 4e2x dx = 4 3 e2x 2 4 –1 —1 2 = 2 3e2x 4 –1 —1 2 = 2(e – e–2) = 21e – 1 e2 2 Example 25 Evaluate ∫ 1 e x2 lnx dx Solution: Put u = ln x ⇒ du dx = 1 x and dv dx = x2 ⇒ v = x3 3 Substituting into ∫ a b u dv dx dx = 3uv4 a b – ∫ a b v du dx dx gives ∫ 1 e x2 lnx dx = 3ln x 1 x3 3 24 1 e – ∫ 1 e x3 3 · 1 x dx = 1 3 3x3 ln x4 1 e – 1 3 3 x3 3 4 1 e = 1 3 (e3 – 0) – 1 9 (e3 – 1) = 2 9 e3 + 1 9 = 1 9 (2e3 + 1)
107 Mathematics Semester 2 STPM Chapter 3 Integration 3 It is important to remember that the term uv on the right-hand side of the expression ∫ u dv dx dx = uv – ∫ v du dx dx has already been integrated and so should be written in square brackets with the limits indicated. Example 26 Evaluate ∫ 2 3 x2 (x – 1)(x + 2) dx Solution: x2 (x – 1)(x + 2) ≡ 1 – x – 2 (x – 1)(x + 2) Let x – 2 (x – 1)(x + 2) ≡ A x – 1 + B x + 2 x – 2 ≡ A(x + 2) + B(x – 1) Let x = 1, –1 = 3A ⇒ A = – 1 3 Let x = –2, –4 = –3B ⇒ B = 4 3 ∫ 2 3 x – 2 (x – 1)(x + 2) dx = ∫ 2 3 31 + 1 3(x – 1) – 4 3(x + 2)4 dx = 3x + 1 3 ln |x – 1| – 4 3 ln |x + 2|4 2 3 = 13 + 1 3 ln 2 – 4 3 ln 52 – 12 + 0 – 4 3 ln 42 = 1 + 1 3 ln 2 + 4 3 ln 4 5 Example 27 By using the substitution u = 2x – 1, evaluate ∫ 0 1 x(2x – 1)8 dx. Solution: Let u = 2x – 1 du dx = 2 ⇒ dx du = 1 2 When x = 0, u = –1 When x = 1, u = 1 Hence, ∫ 0 1 x(2x – 1)8 dx = ∫ –1 1 u + 1 2 · u8 · 1 2 du = 1 4 ∫ –1 1 (u9 + u8 ) du = 1 4 3 u10 10 + u9 9 4 –1 1 = 1 4 31 1 10 + 1 9 2 – 1 1 10 – 1 9 24 = 1 18
108 Mathematics Semester 2 STPM Chapter 3 Integration 3 Note that the limits of integration, originally values of x, are transformed into values of u, thus allowing direct calculation of the required definite integral. Example 28 Evaluate ∫ 0 1 x(1 – x2 ) —3 2 dx by means of the substitution x = sin q. Solution: Let x = sin q dx dq = cos q When x = 0, q = 0 When x = 1, q = π 2 Hence, ∫ 0 1 x(1 – x2 ) —3 2 dx = ∫ 0 —π 2 sin q cos3 q dq = ∫ 0 —π 2 cos4 q sin q dq = –∫ 0 —π 4 cos4 q (–sin q) dq = –3 cos5 q 5 4 0 —π 4 = –10 – 1 5 2 = 1 5 Exercise 3.7 1. Evaluate the following definite integrals. (a) ∫ 3 5 (3x2 + 2x) dx (b) ∫ 0 —2 3 (3x – 2)3 dx (c) ∫ 0 1 1 2x + 1 dx (d) ∫ 1 2 1 1 x – 2 2 2 dx (e) ∫ 0 —1 2 (2 – e–2x ) 2 dx (f) ∫ 0 1 x + 1 (x + 2)(x + 3) dx (g) ∫ 0 1 2x2 + 4x + 5 x2 + 2x + 1 dx (h) ∫ 0 1 x2 e4x dx (i) ∫ 0 π x2 sin x dx (j) ∫ 1 2 x ex dx (k) ∫ 0 —π 2 sin 2x 1 + cos2 x dx
109 Mathematics Semester 2 STPM Chapter 3 Integration 3 2. Evaluate the following integrals using the suggested substitution. (a) ∫ 3 4 x(x – 3)7 dx ; u = x – 3 (b) ∫ 0 1 8 3 + 4x dx ; u2 = 3 + 4x (c) ∫ 0 2 x 4 – x2 dx ; u2 = 4 – x2 (d) ∫ 0 π sin x cos4 x dx ; cos x = u (e) ∫ 0 —π 6 cos q 1 – 2 sin q dq ; x = 1 – 2 sin q (f) ∫ —1 4 — 3 4 1 1 + 16x2 dx ; 4x = tan q 3. (a) Show that d dx 1 x3 1 + x4 2 = x2 (3 – x4 ) (1 + x4 ) 2 . (b) Evaluate ∫ 1 2 x2 (3 – x4 ) (1 + x4 ) 2 dx. 4. (a) Show that ∫ 0 2 x e2x dx = 1 4 (3e4 + 1) (b) Find the exact value of ∫ 0 2 x2 e2x dx. 5. Express x – 1 2x2 + 3x + 1 in partial fractions. Hence, show that ∫ 0 4 x – 1 2x2 + 3x + 1 dx = ln 25 27 . 6. Differentiate e2x (2x – 1) 4 with respect to x. Hence, find the value of ∫ 0 1 x e2x dx. Properties of definite integrals (i) ∫ a b k[f(x)] dx = k ∫ a b f(x) dx, where k is a constant. (ii) ∫ a b [f(x) ± g(x)] dx = ∫ a b f(x) dx ± ∫ a b g(x) dx (iii) ∫ a b f(x) dx = ∫ a c f(x) dx + ∫ c b f(x) dx Applications of integration Area under a curve We have seen that the area of a region bounded by the curve y = f(x), the x-axis and the ordinates x = a and x = b is given by A = ∫ a b f(x) dx.
110 Mathematics Semester 2 STPM Chapter 3 Integration 3 Example 29 Calculate the area of the region bounded by the curve y = 3x2 + 14x + 15, the x-axis, x = –1 and x = 2. Solution: y x A –1 0 2 y = 3x 2 + 14x + 15 Area, A = ∫ –1 2 y dx = ∫ –1 2 (3x2 + 14x + 15) dx = 3x3 + 7x2 + 15x4 –1 2 = (8 + 28 + 30) – (–1 + 7 – 15) = 75 units2 Area below the x-axis When a curve extends below the x-axis, the corresponding y value is negative and so ∫ a b y dx is a negative value. Thus, when an integral turns out to be negative, it indicates that the area is below the x-axis. y = f(x) y x a A b 0 Figure 3.3 A = ∫ a b y dx = –∫ a b f(x) dx
111 Mathematics Semester 2 STPM Chapter 3 Integration 3 Area partly above and partly below the x-axis y = f(x) y ba c x 0 – + Figure 3.4 Integration from x = a to x = c i.e. ∫ a c f(x) dx gives the algebraic sum of the two parts. Hence, it is always wise to draw a sketch of the curve to see whether the curve extends below the x-axis before carrying out the integration. The required area is then calculated by finding the positive and negative parts separately and adding the numerical values of each part to obtain the required area. i.e. A = ∫ a b f(x) dx + ∫ a c f(x) dx. Example 30 Calculate the area of the region bounded by the curve y = x2 – 4, the x-axis, x = –1 and x = 4. Solution: y A B x 0 –1 –2 2 4 y = x 2 – 4 The first step is to draw a sketch of the curve. This shows that the y values are positive for 2 x 4 and negative for –1 x 2. We therefore need to calculate the area in two parts. Area A = ∫ –1 2 (x2 – 4) dx = 3 x3 3 – 4x4 –1 2 = 1 8 3 – 82 – 1– 1 3 + 42 = |–9| = 9 units2 .
112 Mathematics Semester 2 STPM Chapter 3 Integration 3 Area B = ∫ 2 4 (x2 – 4) dx = 3 x3 3 – 4x4 2 4 = 1 64 3 – 162 – 1 8 3 – 82 = 10 2 3 units2 Total area = 9 + 10 2 3 = 19 2 3 units2 If we integrated right through in one calculation, we would have obtained Area = ∫ –1 4 f(x) dx = ∫ –1 4 (x2 – 4) dx = 3 x3 3 – 4x4 –1 4 = 1 64 3 – 162 – 1– 1 3 + 42 = 1 2 3 units2 However, it only gives the correct value of the definite integral and not the correct value of the total area enclosed. Area bounded by two curves y = f(x) y = g(x) y a b A x 0 Figure 3.5 Suppose that we want to find the area of the region bounded by the curves y = f(x) and y = g(x) and suppose that the curves intersect at x = a and x = b. Area A can be treated as the difference between the two areas B and C, as shown in Figure 3.6 and Figure 3.7.
113 Mathematics Semester 2 STPM Chapter 3 Integration 3 y = f(x) y a b B x 0 y = g(x) y a b C x 0 Figure 3.6 Figure 3.7 A = B – C = ∫ a b f(x) dx – ∫ a b g(x) dx = ∫ a b [f(x) – g(x)] dx, where f(x) g(x) for a x b. Example 31 Calculate the area of the region enclosed by the line y = x + 1 and the curve y = x2 – 2x + 1. Solution: First, draw a sketch showing where these graphs intersect. y A x 0 1 1 y = x + 1 y = x 2 – 2x –1 When they intersect, x2 – 2x + 1 = x + 1 x2 – 3x = 0 x(x – 3) = 0 x = 0 or 3 The enclosed area, A = ∫ 0 3 (x + 1) dx – ∫ 0 3 (x2 – 2x + 1) dx = ∫ 0 3 [x + 1 – (x2 – 2x + 1)] dx = ∫ 0 3 (–x2 + 3x) dx = 3– x3 3 + 3 2 x2 4 0 3 = 1–9 + 27 2 2 = 9 2 units2
114 Mathematics Semester 2 STPM Chapter 3 Integration 3 Area between a curve and the y-axis So far, we have calculated areas between curves and the x-axis. We can also use integration to calculate the area between a curve and the y-axis. In such cases, the integral involves dy and not dx. Hence, it is necessary to express x in terms of y wherever it appears. y a A b x 0 y = f(x) Figure 3.8 Area between the curve y = f(x) and the lines y = a and y = b is given by A = ∫ a b x dy Example 32 Calculate the area of the region bounded by the curve y = x – 2 , the y-axis and the lines y = 0 and y = 2. Solution: y x 0 2 y = fix – 2 Area = ∫ 0 2 x dy = ∫ 0 2 (y2 + 2) dy = 3 y3 3 + 2y4 0 2 = 8 3 + 4 = 6 2 3 units2 .
115 Mathematics Semester 2 STPM Chapter 3 Integration 3 Exercise 3.8 1. Calculate the area of the region bounded by the curve y = f(x), the x-axis and the stated ordinates in the following cases. (a) y = x3 + 2x2 + x + 1, x = –1 and x = 2 (b) y = x2 – 25, x = –5 and x = 5 (c) y = –3x2 – 12x + 10, x = 1 and x = 4 2. Sketch the curve and find the area between the curve and the x-axis between the given boundaries. (a) y = x5 – 2 between x = –1 and x = 0 (b) y = x4 – x2 between x = –1 and x = 1 (c) y = x3 + x2 – 2x between x = –3 and x = 2 (d) y = x2 – x – 2 between x = –2 and 3. 3. In each of the following cases, calculate the area enclosed by the given boundaries. (a) y = 10 – x2 and y = x2 + 2 (b) y = 4e2x , y = 4e–x , x = 1 and x = 3 (c) y = x2 – 4x + 20, y = 3x, x = 0 and x = 4 4. Calculate the area enclosed by the curve y = x3 and the straight line y = x. 5. Sketch the curves with equations y = x2 and y = 4x – x2 . Find the coordinates of the points of intersection of the curves. Calculate the area of the region enclosed by the curves. 6. Sketch the curve y = x(4 – x) and the line y = 2x – 3. Find the coordinates of the points of intersection of the line and the curve. Calculate the area of the region enclosed by the line and the curve. 7. Sketch, on the same diagram, the curves given by the equations xy = 6 and y = 9 – 3x, x 0 Show, by integration, that the area bounded by the two curves is 9 2 – 6 ln2. 8. Find the coordinates of the point of intersection of the curves y = ex and y = 2 + 3e–x . If both curves cut the y-axis at the points A and B, calculate the area bounded by the line AB and the arcs AC and BC. 9. The curves y = 3 sin x and y = 4 cos x 10 < x < π 2 2 intersect at point A and meet the x-axis at the origin O and point B1 π 2 , 02 respectively. Prove that the area enclosed by the arcs OA, AB and the line OB is 2 square units. Volumes of solids of revolution If the plane figure bounded by the curve y = f(x), the x-axis and the ordinates x = a and x = b is revolved completely about the x-axis, it will generate a solid symmetrical about Ox, called a solid of revolution.
116 Mathematics Semester 2 STPM Chapter 3 Integration 3 y = f(x) O y x a b y = f(x) O V y x Figure 3.9 Figure 3.10 Let V be the volume of the solid generated. To find V, let us first consider a thin strip of the original plane figure. y = f(x) O x y P Q y δx δx δv y O y x Figure 3.11 Figure 3.12 The volume generated by the strip is approximately equal to the volume generated by the rectangle which is approximately cylindrical with radius y and thickness x. Its volume is given by V ≈ πy2 dx. If we divide the whole plane figure into a number of such strips, the volume of the solid is the sum of all these discs as x → 0. Total volume, V ≈ x = b x Σ = a πy2 dx = lim dx → 0 x = b x Σ = a πy2 dx = ∫ a b πy2 dx Hence, volume of the solid of revolution = π∫ a b y2 dx When a region is revolved about the y-axis, the volume of the solid of revolution is given by V = π∫ a b x 2 dy In this case, you will need to substitute x2 in terms of y. y = f(x) a b O y x Figure 3.13
117 Mathematics Semester 2 STPM Chapter 3 Integration 3 Example 33 The region between the curve y = x2 , the x-axis and the lines x = 1 and x = 3 is revolved completely about the x-axis. Find the volume of the solid of revolution. Solution: y x 0 1 3 y = x 2 The region is shaded in the diagram. Volume generated = π∫ 1 3 y2 dx = π∫ 1 3 x4 dx = π3 x5 5 4 1 3 = π1 243 5 – 1 5 2 = 242 5 π units3 . Example 34 Using integration, find the volume of a cylindrical ball of radius 2 cm. Solution: The volume is obtained by revolving the top half of the circle x2 + y2 = 4 about the x-axis. y x x 2 + y 2 = 4 2 0 2 –2 –2 Volume of spherical ball = π∫ 2 –2 y2 dx = π ∫ 2 –2 (4 – x2 ) dx = π34x – x3 3 4 2 –2 = π318 – 8 3 2 – 1–8 + 8 3 24 = 32 3 π cm3 .
118 Mathematics Semester 2 STPM Chapter 3 Integration 3 Example 35 The region between the curve y = x2 – 1, the y-axis and the lines y = 2 and y = 5 is revolved completely about the y-axis. Find the volume of the solid of revolution. Solution: Volume generated = π∫ 2 5 x2 dy = π∫ 2 5 (y + 1) dy = π 3 y2 2 + y4 2 5 = π3125 2 + 52 – (2 + 2)4 = 27 2 π units3 . Volume of a solid of revolution generated by a region between two curves y = f(x) y = g(x) y a b R x 0 Figure 3.14 If f(x) g(x) for a x b and R is the region bounded by the curves y = f(x) and y = g(x), and the lines x = a and x = b, the volume of revolution when R is revolved about the x-axis is given by V = π ∫ a b {[f(x)]2 – [g(x)]2 } dx Example 36 The region bounded by the curves y = x2 and y = x is revolved completely about the x-axis. Find the volume of the solid formed. Solution: Volume generated = π∫ 0 1 (x2 – x4 ) dx = π3 x3 3 – x5 5 4 0 1 = π1 1 3 – 1 5 2 = 2 15 π cubic units y x 0 y = x 2 – 1 –1 2 5 y x 0 y = x 2 y = x
119 Mathematics Semester 2 STPM Chapter 3 Integration 3 Exercise 3.9 1. Calculate the volume of the solid of revolution when the bounded region is revolved completely about the x-axis. (a) y = 2x, the x-axis and the lines x = 1 and x = 3 (b) y = x2 + 1, the x-axis and the lines x = –1 and x = 1 (c) y = x , the x-axis and the line x = 4 2. Calculate the volume of the solid of revolution when the bounded region is revolved completely about the y-axis. (a) y = 3x, the y-axis and the lines y = 3 and y = 6 (b) y = x – 3, the y-axis, the x-axis and the line y = 6 (c) y = x2 – 2, the y-axis and the line y = 4. 3. The region bounded by the two curves is revolved completely about the x-axis. Calculate the volume of the solid formed. (a) y = x(6 – x) and y = 3x (b) y2 = 4x and y = 2x (c) y2 = 4x and x2 = 4y. 4. The region R in the first quadrant is bounded by the y-axis, the x-axis, the line x = 3 and the curve y2 = 4 – x. Calculate the volume of the solid formed when R is revolved about the y-axis. 5. A hemispherical bowl is formed by revolving the bottom half of the circle x2 + y2 = 100 about the y-axis. (a) Find the volume of the bowl. (b) If the bowl is filled with water to a depth of 8 cm, find the volume of the water in the bowl. 6. Sketch, on the same axes, the curve y = 16 – x2 and the line y = 6x in the first quadrant. The region bounded by the curve and the line is revolved completely about the x-axis. Find the volume of the solid generated, leaving your answer as a multiple of π. 7. The area of the region bounded by the curve y = tan x, the x-axis and the line x = π 3 is revolved about the x-axis. Calculate the volume of the solid formed. 8. Calculate the volume of the solid generated when the finite region enclosed by the curve y = 1 + 2e–x and the lines x = 0, x = 1 and y = 1 is revolved completely about the x-axis. 9. Sketch the curve y = ex and y = e–x for –2 x 2. The interior of a wine glass is formed by revolving the curve y = ex from x = 0 to x = 2 about the y-axis. If the units are in centimetres, find, correct to 2 significant figures, the volume of the wine that the glass contains when full. 10. Sketch the curve whose equation is y = 1 – 1 x + 2 , indicating any asymptotes. The region bounded by the curve, the x-axis and the lines x = 0 and x = 2 is denoted by R. Find the volume of the solid formed when R is revolved about the x-axis.
120 Mathematics Semester 2 STPM Chapter 3 Integration 3 11. The diagram shows the region R in the first quadrant bounded by the curves y = 1 4 (4 – x2 ), y = 1 2 (4 – x2 ) and the y-axis. Calculate the volume of the solid formed when R is revolved about the y-axis. 12. A chord of the circle x2 + y2 = r 2 is parallel to the x-axis and of length 2l. The minor segment cut off by this chord is revolved about the x-axis to form a solid of revolution. Show that its volume is 4 3 π l 3 . Summary 1. If d dx [F(x)] = f(x), then F(x) + c is an anti-derivative of f(x), where c is an arbitrary constant. The indefinite integral ∫f(x) dx is defined by ∫f(x) dx = F(x) + c. 2. Standard integrals f(x) ∫f(x) dx xn xn + 1 n + 1 + c, n ≠ –1 1 x ln |x| + c ex ex + c sin x –cos x + c cos x sin x + c sec2 x tan x + c 3. Integration by substitution (i) ∫ (ax + b) n dx = (ax + b) n + 1 a(n + 1) + c, n ≠ –1 (ii) ∫ 1 ax + b dx = 1 a ln |ax + b| + c (iii) ∫ [f(x)]n f(x) dx = [f(x)]n + 1 n + 1 + c, n is a real number (iv) ∫ f(x) f(x) dx = ln |f(x)| + c (v) ∫ eax + b dx = 1 a eax + b + c (vi) ∫sin (ax + b) dx = – 1 a cos (ax + b) + c (vii) ∫ cos (ax + b) dx = 1 a sin (ax + b) + c (viii) ∫sec2 (ax + b) dx = 1 a tan (ax + b) + c y R x 0 1 2
121 Mathematics Semester 2 STPM Chapter 3 Integration 3 4. Integration by parts If u and v are functions of x, ∫ u dv dx dx = uv – ∫ v du dx dx 5. The definite integral of f from a to b is defined by ∫ a b f(x) dx = lim n → ∞ n i Σ = 1 f(xi ) dx, where dx = 1 n (a – b) and xi is any value of x in the ith interval. 6. Properties of definite integrals (i) ∫ a b k[f(x)] dx = k ∫ a b f(x) dx, where k is a constant. (ii) ∫ a b [f(x) ± g(x)] dx = ∫ a b f(x) dx ± ∫ a b g(x) dx (iii) ∫ a b f(x) dx = ∫ a c f(x) dx + ∫ c b f(x) dx 7. Integration is a process of summation ∫ a b f(x) dx = lim dx → 0 x = b x Σ = a f(x) dx 8. Plane areas y x a A 0 b y = f(x) y a B b x 0 y = f(x) Area A = ∫ a b y dx Area B = ∫ a b x dy = ∫ a b f(x) dx y C a b x 0 y = f(x) y = g(x) Area C = ∫ a b [f(x) – g(x)] dx
122 Mathematics Semester 2 STPM Chapter 3 Integration 3 9. Volumes of solids of revolution a b y x 0 y a b x 0 About the x-axis, V = π∫ a b y2 dx About the y-axis, V = π∫ a b x2 dy STPM PRACTICE 3 1. Find the following indefinite integrals. (a) ∫ 1 2x – 9 dx (b) ∫ 2x + 3 3x2 + 9x – 1 dx (c) ∫ e3x + 4 e2x dx (d) ∫ x 2x – 1 dx (e) ∫ x 4 – x2 dx (f) ∫ x2 e–x3 dx (g) ∫ x + 1 x(2x + 1) dx (h) ∫ x(2x + 1) x + 1 dx 2. Evaluate (a) ∫ 2 3 1 x(x2 – 1) dx (b) ∫ 0 1 x (1 – x) —1 2 dx (c) ∫ 0 1 x e–3x dx (d) ∫ 0 1 x 1 + x dx (e) ∫ 1 e (2x + 1) ln x dx (f) ∫ 0 3 x 1 + x2 dx (g) ∫ 1 4 1 3 x – x 2 2 dx (h) ∫ 0 1 1 – 4x 3 + x – 2x2 dx 3. Find the value of ∫ 0 1 (2 + x) ln(x + 2) dx. 4. Using the substitution x = 5 sin q, find ∫ dx x2 25 – x 2 . 5. Show that ∫ x2 cos–1 x dx = x3 3 cos–1 x – 1 9 (2 + x2 ) 1 – x 2 + c. 6. Find d dx (x cos x). Hence, evaluate (a) ∫ 0 π cos x dx – ∫ 0 π x sin x dx (b) ∫ 2 π x sin x dx 7. Show that ∫ x2 tan–1 x dx = x3 3 tan–1 x – x2 6 + 1 6 ln(1 + x2 ) + c.
123 Mathematics Semester 2 STPM Chapter 3 Integration 3 8. By using the substitution u = sin-1 x, find the exact value of ∫ √3 2 2 sin–1 x 1 – x2 dx. 9. Evaluate. (a) ∫ 0 1 8 3 + 4x dx (b) ∫ 0 1 8 (3 + 4x) dx (c) ∫ 0 1 8x 3 + 4x dx 10. (a) Show that ∫ 1 2 (x – 1)(5x + 2) (2x – 1)(x2 + 2) dx = 1 2 ln 8 3 . (b) Using the substitution x = 1 2 (1 + sin q), show that ∫ —1 4 —3 4 x x – x 2 dx = 1 2 ∫ —π 6 –—π 6 (1 + sin q) dq. Hence, evaluate the integral. 11. (a) Show that ∫ 0 1 x2 ex dx = e – 2. (b) Prove that ∫ 0 —π 2 x cos x dx = π 2 – 1. 12. Express 1 1 – x2 in partial fractions. Hence, show that, if –1 x 1, ∫ 1 1 – x2 dx = 1 2 ln 1 1 + x 1 – x 2 + c, where c is the constant of integration. Using integration by parts, show that ∫ 1 1 – x2 dx = x 1 – x2 – ∫ 2x2 (1 – x2 ) 2 dx. Deduce the value of ∫ 0 —1 2 x2 (1 – x2 ) 2 dx correct to three significant figures. 13. The graph of y = (x – 1)4 – 1 is as shown below. y A B x –1 0 2 y = (x – 1)4 – 1 Calculate the total area of the shaded regions A and B. 14. Sketch the curves y = (x – 2)2 + 1 and y = 6 – (x – 3)2 . Find the coordinates of their points of intersection. Show that the area enclosed by the two arcs between their points of intersection is 9. 15. Calculate the area of the region in the first quadrant bounded by the curve y = x2 + 4, the line y = 8 and the y-axis. This region is revolved about the y-axis. Calculate the volume of the solid of revolution formed.
124 Mathematics Semester 2 STPM Chapter 3 Integration 3 16. The diagram shows the curve y = x2 – 4. The region R is bounded by the line y = 12, the x-axis, the y-axis and the curve y = x2 – 4 for positive values of x. The inside of a vase is formed by revolving R about the y-axis. Each unit of x and y represents 2 cm. (a) Write down an expression for the volume of revolution about the y-axis. (b) Find the capacity of the vase in litres. (c) Show that if the vase is filled to 5 6 of its internal height, it is three-quarters full. 17. A mathematical model for a large garden pot is obtained by revolving about the y-axis the part of the curve y = 0.1x2 which is between x = 10 and x = 25 and then adding a flat base. Units are in centimetres. (a) Sketch the curve and shade the cross-section of the pot, indicating which line will form its base. (b) Garden compost is sold in litres. Find the number of litres required to fill the pot to a depth of 45 cm. (Ignore the thickness of the pot). 18. Find the coordinates of A and B, the points of intersection of the circle x2 + y2 = 25 and the line y = 4 as shown in the diagram below. 0 4 A B y x y = 4 x 2 + y 2 = 25 A napkin ring is formed by revolving the shaded region about the x-axis. Find the volume of the napkin ring. 19. The region bounded by the lines x = 0, x = 1, y = 0 and the curve y = 1 2 – x is denoted by R. Calculate the area of R and the volume of the solid of revolution formed when R is revolved about the x-axis. 20. Region R which is bounded by the curve y = x 1 + x2 , line y = x and vertical line x = 3 is shown in the diagram below. y = x 0 y x R fiffff y = ––––– 1 + x2 fiff3 x (a) Calculate the area of region R. (b) Calculate the volume of the solid generated when the region R is rotated completely about the x-axis. y R x –2 0 2 –4 12
125 Mathematics Semester 2 STPM Chapter 3 Integration 3 21. The equation x2 + y2 = 1 represents a circle with centre at the origin and radius 1 unit. By considering an appropriate region of the circle, show that, ∫ 0 1 1 + x2 dx = π 4 . The diagram shows a circle with equation x2 + (y – 1)2 = 1. The region R is bounded by the circle, the x-axis and the line x = 1. Show that the volume of the solid formed when R is revolved about the x-axis is given by π ∫ 0 1 (2 – x2 – 21 – x2 ) dx, and find this volume, giving your answer in terms of π. 22. Sketch the curve y = 1 + 2e–x , showing clearly the behaviour of the curve as x → ∞. Find the area of the finite region enclosed by the curve and the lines x = 0, x = 1 and y = 1. Find the volume of the solid formed when this region is revolved completely about the line y = 1. 23. Express 8 + 2x – x2 in the form of P – (x – Q) 2 where P and Q are constants. Hence, evaluate ∫ 0 —3 2 dx 8 + 2x – x 2 . 24. Sketch the arc of the curve y = 2x – x2 for which y is positive. Calculate the area of the region which lies between this curve and the x-axis. If this region is revolved completely about the x-axis, calculate the volume of the solid of revolution generated. 25. Show that ∫ 0 1 10x + 2 (1 + 3x)(1 + 3x2 ) = 1 3 (√3π – ln 2). 26. The line y = 3x + 1 meets the curve y = x2 + 3 at the points A and B. (a) Sketch, on the same coordinate axes, the curve and the straight line. (b) Determine the coordinates of A and B. (c) Calculate the area of the region R bounded by the curve and the straight line. (d) Find the volume of the solid formed when R is rotated through 360° about the x-axis. 27. Using the substitution x = 2 sin q, find ∫ 0 1 x2 4 – x2 dx correct to four decimal places. 28. Express 9x2 + 12x + 7 in the form (ax + b) 2 + c where a, b and c are constants whose values are to be found. Hence, find ∫ 1 9x2 + 12x + 7 dx. 29. Show that ∫ 0 1 1 1 + x + x2 dx = 1 9 π√3 . 30. The equations of two curves are y = x2 + 2 and y = 6 – x2 . (a) Sketch the two curves on the same coordinate axes. (b) Find the coordinates of the points of intersection of the two curves. (c) Calculate the volume of the solid formed when the region bounded by the two curves is revolved completely about the y-axis. O 1 1 y x R
126 CHAPTER DIFFERENTIAL 4 EQUATIONS Learning Outcome (a) Find the general solution of a first order differential equation with separable variables. (b) Find the general solution of a first order linear differential equation by means of an integrating factor. (c) Transform, by a given substitution, a first order differential equation into one with separable variables or one which is linear. (d) Use a boundary condition to find a particular solution. (e) Solve problems, related to science and technology, that can be modelled by differential equations. boundary condition – syarat sempadan differential equation – persamaan pembezaan first order – tertib pertama general solution – penyelesaian am integrating factor – faktor pengamir linear – linear particular solution – penyelesaian khusus separable variables – pemboleh ubah terpisahkan substitution – gantian transform – jelmakan variable – pemboleh ubah Bilingual Keywords
127 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 4.1 Differential Equations Differential equations Consider x as an independent variable and y as a dependent variable. An equation that involves at least one derivative of y with respect to x, e.g. dy dx , d2 y dx 2 , …, dn y dxn is known as a differential equation. A few examples of differential equations are as follows. (a) dy dx + xy2 = sin x (b) 2 d2 y dx 2 – 3 dy dx + 4y = x (c) (1 + x2 ) 1 dy dx 2 2 + 3y = 0 Order of differential equation The order of a differential equation is the highest order of the derivative found in the differential equation. In the above examples, equations (a) and (c) are differential equations of the first order because the equations consist of only derivatives of the first order, dy dx . Equation (b) is a differential equation of the second order since it involves a second order derivative, d2 y dx 2 . Degree of differential equation The degree of a differential equation is determined by the power of the highest order of the derivative in the equation. Hence, equations (a) and (b) are differential equations of the first degree since no derivatives are raised to any power except one. Equation (c) is of second degree as the highest derivative dy dx is of power 2. In this chapter, only differential equations of the first order and first degree that can be solved by separating the variables or that can be transformed into such equations will be dealt with in further details. Example 1 Determine the order and the degree of the following differential equations. (a) dy dx + y = x2 (b) x2 (1 + y) 1 dy dx 2 2 – (1 + x)y2 = 0 (c) d2 y dx 2 = 11 + dy dx 2 —3 2 Differential Equation INFO Introduction to Differential Equation VIDEO
128 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 Solution: (a) order : 1, degree : 1 (b) order : 1, degree : 2 (c) order : 2 From d2 y dx2 = 11 + dy dx 2 —3 2 1 d2 y dx2 2 2 = 311 + dy dx 2 —3 2 4 2 1 d2 y dx2 2 2 = 11 + dy dx 2 3 Thus, the degree is 2. Solution of differential equations The function y = φ(x) is the solution of a differential equation if it satisfies a given differential equation. Example 2 Show that y = 1 2 x2 + C is the solution of the differential equation dy dx = x. Solution: From y = 1 2 x2 + C, dy dx = x Thus, y = 1 2 x2 + C is the solution of the differential equation dy dx = x. Example 3 Show that y = Aex + (x + 2)e2x is a solution of the differential equation dy dx – y = (x + 3)e2x . Solution: From y = Aex + (x + 2)e2x , dy dx = Aex + e2x + 2(x + 2)e2x = Aex + (2x + 5)e2x Substitute this into the given equation, dy dx – y = Aex + (2x + 5)e2x – [ Aex + (x + 2)e2x ] = (x + 3)e2x Thus, y = Aex + (x + 2)e2x is a solution of the differential equation dy dx – y = (x + 3)e2x . From these two examples, note that the solution of the differential equation of the first order will contain one arbitrary constant.
129 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 A solution of the differential equation that contains an arbitrary constant is known as a general solution. Example 4 Show that y2 = A1 x – 1 x + 12 is a solution of the differential equation dy dx = y x2 – 1 . Hence, find the value of A if y = 1 when x = 2. Solution: Given y2 = A1 x – 1 x + 12 ................................... Thus, 2y dy dx = 2A (x + 1)2 ...................................... From , A = y2 1 x + 1 x – 1 2 Substitute into equation , 2y dy dx = 2y2 (x + 1)2 1 x + 1 x – 1 2 i.e. dy dx = y x2 – 1 Thus, y2 = A1 x – 1 x + 12 is a solution of the differential equation dy dx = y x2 – 1 . Hence, with the condition y = 1 when x = 2, from equation , 12 = A1 2 – 1 2 + 12 A = 3 Thus, y2 = 3 1 x – 1 x + 12 In Example 4, the value of A can be determined from the given condition. A general solution that satisfies certain conditions (either initial conditions or boundary conditions) so that the value of the arbitrary constant in the solution is obtained is called the particular solution. Family of curves of the solution The equation y = 1 2 x2 + C represents a family of curves with similar characteristics. Each value of C will give rise to a particular curve in the family. Figure 4.1 shows sketches of four members of the family curves for the solution corresponding to C = –1, C = 0, C = 1 and C = 2. Figure 4.1 y x C = 2 C= 1 C= 0 C= –1 0
130 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 Exercise 4.1 1. State the order and degree of the following differential equations. (a) x2 d2 y dx2 + x dy dx + 4y = 0 (b) 2x d2 y dx2 + (sin x) dy dx – y = cos x (c) 1 dy dx 2 2 + x = y (d) d2 y dx2 – x = dy dx (e) 1 d3 y dx3 2 2 + x d2 y dx2 + 1 dy dx 2 5 + y4 = cos x 2. State the degree of the following differential equations. (a) dy dx = 3 1 + 1 d2 y dx2 2 2 4 —5 3 (b) 1y – x dy dx 2 2 = 1 + 1 dy dx 2 2 (c) d2 s dt 2 = 5 + 1 ds dt 2 2 3. Show that y = 3x + 2A x is the general solution of the differential equation x dy dx + y = 6x. Hence, obtain the particular solution of the differential equation x dy dx + y = 6x if y = 1 when x = 1. 4. Show that y + 2 = Ax + x2 is the general solution of the differential equation x dy dx – y = 2 + x2 . Hence, obtain the particular solution of the differential equation x dy dx – y = 2 + x2 , given that y = 1 when x = 1. 4.2 First Order Differential Equations with Separable Variables Generally, a first order differential equation can be written as dy dx = f(x, y)..................................................... In certain cases, function f(x, y) can be written as f(x, y) = u(x) . v(y) .............................................. where u(x) and v(y) are functions of x and y respectively. By substituting equation into equation , dy dx = u(x) . v(y) Equation is a differential equation with separable variables if it can be written as dy v(y) = u(x) dx
131 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 Example 5 Determine whether each of the following differential equations is a differential equation with separable variables. (a) dy dx = y + 3x2 y (b) dy dx = x + y x (c) dy dx = e x + y Solution: (a) From dy dx = y + 3x2 y, dy dx = y(1 + 3x2 ) Rearranging dy y = (1 + 3x2 ) dx dy v(y) = u(x) dx Thus, dy dx = y + 3x2 y is a differential equation with separable variables. (b) dy dx = x + y x is not a differential equation with separable variables. (c) From dy dx = ex + y dy dx = ex ey dy ey = ex dx Thus, dy dx = ex + y is a differential equation with separable variables. To find the solution of a differential equation with separable variables, integrate the equation dy v(y) = u(x) dx separately. The result of the integration is ∫ dy v(y) + C1 = ∫ u(x) dx + C2 where C1 and C2 are integral constants. Rearrange it, ∫ dy v(y) = ∫ u(x) dx + C2 – C1 = ∫ u(x) dx + C C = C2 – C1 This seems to be the solution of the first order differential equation with separable variables.
132 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 Example 6 Find the general solution of the differential equation dy dx = 1 + y2 2y . Hence, find the particular solution if y = 0 when x = 0. Solution: From dy dx = 1 + y2 2y , 2y 1 + y2 dy = dx ∫ 2y 1 + y2 = ∫ dx ln |1 + y2 | = x + C where C is an arbitrary constant 1 + y2 = ex + C y2 = Aex – 1 where A = eC Thus, the general solution of the differential equation is y = ±Aex – 1 . When x = 0, y = 0, C = 0 and A = 1. Therefore, the particular solution of the differential equation is y = ±ex – 1 . The particular solution can also be obtained by using definite integration with the initial condition as the lower limit and x and y as the upper limit. Example 7 The gradient function of a curve at any point (x, y) is given by the equation dy dx = 2y 1 – x2 . Find the equation of the curve if the curve passes through the point 1 1 2 , 12. Solution: The equation of the gradient of the curve can be written as dy y = 2 dx 1 – x2 ∫ dy y = ∫ 2 dx 1 – x2 ln | y | = ln 1 + x 1 – x + C When x = 1 2 and y = 1, C = – ln 3 i.e. ln | y | = ln 1 + x 1 – x – ln 3 Thus, the equation of the curve that passes through the point 1 1 2 , 12 is y = 1 + x 3(1 – x) .
133 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 Exercise 4.2 Find the general solution of the following differential equations. 1. dy dx = x2 + 1 x2 2. dr dt = 3 + t 2 t 3. 1 x dy dx = 1 x2 – 1 4. x dx dt = t + 2 5. dy dx = y2 – 4 6. x dy dx = –y2 7. dy dx = 2 + y2 y 8. (x – 2) dy dx – y = 0 9. x dy dx = xy + y 10. dy dx = 1 yex 11. y2 x dy dx = ln x 12. x dy dx = 1 – 2y + y2 13. (y2 + 1) dy dx + 2xy2 = 2x 14. 2y dy dx – xy2 = x Solve the following differential equations. 15. dy dx = – y x 16. dy dx = 1 x – 1 17. dy dx = y + 2 18. (x2 – 2) dy dx – 2xy = 0 19. (x – 1) dy dx – y = 0 20. dy dx = 2xy + y Find the particular solution of the following differential equations with the given initial conditions. 21. y dy dx = x + 2, y = 2 when x = 1 22. dy dx = 1 x + x, y = 1 when x = 1 23. dy dx = 2y – 3, y = 2 when x = 0 24. dy dx = 1 + x x – 1 , y = 1 when x = 2 25. dy dx = 2x 1 + x2 , y = 0 when x = –1 26. dy dx = 3y2 – 1 6y , y = 1 when x = –1 27. e–x dy dx = 1, y = –1 when x = 0 28. xy dy dx = ln x, y = 0 when x = 1 29. dy dx + y2 = 4, y = 0 when x = ln 2 30. ex + y dy dx = x, y = 1 when x = 0 31. y dy dx = 9 – 4y 2 , y = 0 when x = 0 32. dy dx = xx2 + 9 , y = 9 when x = 0 33. dy dx + 1 x = ey x , y = ln 2 when x = 1 34. xy dy dx = x2 – 1 y – 1 , y = 0 when x = 1 35. (1 + x2 ) dy dx + 2xy = 4x, y = 0 when x = 1
134 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 4.3 First Order Linear Differential Equations The first order linear differential equation is of the form dy dx + f(x)y = g(x) where f(x) and g(x) are functions of x. It can be solved by multiplying both sides of the equation with e ∫f(x)dx which is called the integrating factor; dy dx + f(x)y = g(x) 3 dy dx + f(x)y4e ∫f(x)dx = g(x)e ∫f(x)dx 3 dy dx e ∫f(x)dx + 1f(x) e ∫f(x)dx 2y4 = g(x) e ∫f(x)dx Note that the left hand side is the differentiation of ye ∫f(x)dx . i.e. d dx y e ∫f(x)dx = 1 d dx y2e ∫f(x)dx + y1 d dx e ∫f(x)dx 2 = 1 dy dx 2e ∫f(x)dx + y 1f(x) e ∫f(x)dx 2 \ 3 d dx 1ye ∫f(x)dx 24 = g(x)e ∫f(x)dx Integrating both sides with respect to x: ∫1 d dx ye ∫f(x)dx 2 dx = ∫ g(x)e ∫f(x)dx dx \ ye ∫f(x)dx = ∫ g(x)e ∫f(x)dx dx + C \ y = ∫ g(x)e ∫f(x)dx dx e ∫f(x)dx + Ce–∫f(x)dx Example 8 Find the general solution to the differential equation x dy dx + 2x2 = y. Solution: x dy dx + 2x2 = y Arrange the differential equation to the form dy dx + f(x)y = g(x). Divide by x: dy dx + 2x = y x dy dx – 1 1 x 2y = –2x dy dx + 1– 1 x 2y = –2x \ f(x) = – 1 x and g(x) = –2x \ The integrating factor is e ∫ –—1 x dx = e–lnx = elnx–1 = x–1. (Notes: eln a = a)
135 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 Multiplying both sides of the differential equation with the integrating factor is e– 1 ∫ x dx . dy dx 1e –∫ —1 x dx 2 – 1 x 1e –∫ —1 x dx 2 y = –2x 1e –∫ —1 x dx 2 d dx 1ye –∫ —1 x dx 2 = –2xe –∫ —1 x dx \ 1ye –∫ —1 x dx 2 = ∫–2xe –∫ —1 x dx dx \ y(x–1) = ∫– 2x(x–1) dx yx–1 = ∫–2 dx yx–1 = –2x + A \ y = –2x2 + Ax Example 9 Find the general solution to the differential equation dy dx = x – 2y. Solution: dy dx = x – 2y Arrange the differential equation to the form dy dx + f(x)y = g(x). dy dx + 2y = x \ f(x) = 2 and g(x) = x. So the integrating factor is e ∫ 2dx = e2x . Multiplying both sides of the equation with e2x ; dy dx (e2x ) + 2(e2x )y = x(e2x ) \ d dx (ye2x ) = xe2x \ ye2x = ∫ xe2x dx + C Integrating the right side by parts: du dx = e2x and v = x \ u = ∫ e2x dx = 1 2 e2x and dv dx = 1 \ ye2x = 1 2 xe2x – ∫ 1 2 e2x dx + C \ ye2x = 1 2 xe2x – 1 4 e2x + C \ y = 1 2 x – 1 4 + Ce–2x
136 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 Exercise 4.3 1. Solve the differential equations (a) (1 + x2 ) dy dx + xy = x + x3 (b) dy dx + 3y = x (c) x dy dx + y = 2x (d) dy dx = 2x2 – y (e) 3 dy dx + 15y = 24e3x 2. Solve the differential equations given the boundary conditions: (a) x dy dx – 3y = x(1 + x3 ); x = 1, y = 1 (b) (1 + x2 ) dy dx + xy = x ; x = 1, y = 0 (c) dy dx + 3y = 2e–x ; x = 0, y = 1 (d) dy dx – 2xy = x ; x = 1, y = 1 4.4 Transformations Differential Equations Some first order differential equations which are not of the form where the variables are separable can be reduced to the separable forms by suitable substitutions. Example 10 Solve the differential equation dy dx + x2 y = – y x . Solution: The variables of the differential equation dy dx + x2 y = – y x are not separable. Let y = v x Differentiating with respect to x: dy dx = x–1 d dx v + v d dx x–1 dy dx = x–1 dv dx + v(–x–2) \ dy dx = 1 x dv dx – v x2
137 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 So dy dx + x2 y = – y x reduces to 1 1 x dv dx – v x2 2 + x2 1 v x 2 = – 1 x 1 v x 2 1 x dv dx – v x2 + vx = – v x2 1 x dv dx + vx = 0 \ dv dx = –vx2 which is the form where the variables are separable. Separating the variables and integrating with respect to x. \ ∫ dv v = –∫ x2 dx \ ln v = – x3 3 + A Substituting v = yx \ ln yx = – x3 3 + A is the general solution. Example 11 Solve the differential equation dy dx = y3 – x2 y x3 + xy2 . Solution: dy dx = y3 – x2 y x3 + xy2 Substituting y = vx, v + x dv dx = v3 – v 1 + v2 dy dx = v + x dv dx x dv dx = v3 – v 1 + v2 – v x dv dx = – 2v 1 + v2 ∫ 1 + v2 v dv = ∫ – 2 x dx ln | v | + 1 2 v2 = –2 ln | x | + C C is an arbitrary constant. ln | vx2 | + 1 2 v2 = C ln | yx | = – y2 2x2 + C yx = Ae – y2 — 2x2 where A = eC
138 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 Example 12 If v = x + y, show that the equation dy dx = x + y + 5 x + y – 3 can be reduced to dv dx = 2(v + 1) v – 3 , and hence solve the equation. Solution: dy dx = x + y + 5 x + y – 3 Substituting v = x + y or y = v – x, d dx (v – x) = v + 5 v – 3 dv dx – 1 = v + 5 v – 3 dv dx = v + 5 v – 3 + 1 = 2(v + 1) v – 3 ∫ v – 3 v + 1 dv = 2∫ dx ∫11 – 4 v + 1 2 dv = 2∫ dx v – 4 ln | v + 1 | = 2x + C x + y – 4 ln | x + y + 1 | = 2x + C ln | x + y + 1 | = 1 4 (y – x) + D D = – 1 4 C x + y + 1 = Ae —1 4 (y – x) A = eD x + y + 1 – Ae —1 4 (y – x) = 0 Example 13 By writing u = 1 y 2 , reduce the equation 2 dy dx – y x = y3 to du dx + u x = –1. Hence solve the differential equation. Solution: Differentiating u = 1 y2 with respect to x du dx = d dy y–2 = – 2 y3 dy dx Dividing the equation 2 dy dx – y x = y3 with y3 . 2 y3 dy dx – 1 xy2 = 1 \ – du dx – u x = 1 \ du dx + u x = –1 The integrating factor is e 1 ∫ x dx = elnx = x.
139 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 Multiplying both sides with the integrating factor e 1 ∫ x dx : du dx e 1 ∫ x dx + u x e 1 ∫ x dx = –1(e 1 ∫ x dx ) du dx x + u = –x d dx (ux) = –x \ ux = ∫ –x dx + A \ ux = – x2 2 + A Substituting u = 1 y2 , \ x y2 = – x2 2 + A \ y2 = 2x –x2 + 2A \ y2 = ± 2x –x2 + 2A Exercise 4.4 1. Solve the differential equations (a) dy dx = x + y x (b) dy dx = x – 2y x (c) dy dx = x2 + 2y2 xy (d) xy dy dx = x2 + y2 (e) xy dy dx = x2 – y2 (f) dy dx = y3 – yx2 x3 + y2 x 2. Show that, by substituting y = vx, the differential equation dy dx = xy x2 + y2 can be transformed into x dv dx + v3 1 + v2 = 0. Hence, find the particular solution if y = 2 when x = 1. 3. Using the substitution y – x = z, show that the differential equation dy dx = y – x + 1 y – x + 5 can be transformed into dz dx = – 4 z + 5 . Hence, solve the differential equation. 4. Using the substitution z = x + y, solve the differential equation dy dx = x + y + 2 x + y + 1 . 5. Solve the differential equation dy dx = 2x – y – 1 2x – y + 3 by using the substitution v = 2x – y.
140 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 6. Using the substitution y = vx, find the general solution of the differential equation 2xy dy dx = y2 – 4x2 . Show that the particular solution which satisfies the condition y = a (a 0) when x = 1 2 a is y2 = 4x(a – x). 7. Show that the differential equation xy2 dy dx – x3 – y3 = 0 can be transformed into the differential equation x dv dx = 1 v2 by using the substitution y = vx. Hence, find the particular solution of the original equation given that x = 1, y = 1. 8. Show that the substitution v = x3 + y transforms the differential equation (x – 1) dy dx – 3x2 – 3y = 0 into the differential equation dv dx = 3v (x – 1). Hence, find the general solution to the original differential equation. 9. By substituting z = 1 y2 , transform the differential equation dy dx + y = xy3 into a differential equation containing z and x. Solve the equation for z, and, hence, show that the solution of the given differential equation for which y = 2 when x = 0 is y = 1 2 2x + 12 —1 2 . 10. By substituting u = y2 , transform the differential equation 2xy dy dx = y2 – 4x2 into a differential equation containing u and x. Find the particular solution of the given differential equation for which y = a when u = 1 2 a. 11. By writing u = 1 y , reduce the differential equation dy dx + y x = y2 to du dx – u x = –1. Hence, solve the original differential equation when x = 1, y = 1. 4.5 Problems Modelled by Differential Equations There are many physical situations in which different variables changes at different rates. Differential equations always arise when we model these physical situations in mathematics. The following situations show a few physical problems that involve differential equations. (a) Assuming a particle falls from rest in a medium that causes the velocity to increase at a rate proportional to its velocity. By denoting velocity as v and time as t, the rate of increase of velocity can be written as dv dx . Hence, the velocity of the particle satisfies the differential equation dv dt = kv. (b) The rate of decay of a radioactive substance is proportional to the amount of remaining substance. Hence, the amount of remaining substance, x at any time t can be found by solving the differential equation dx dt = –kx. (c) In a chemical process, a certain substance A continuously changes to another substance B. The rate of such a change is proportional to the mass of A and inversely proportional to the mass of B, at any time t. The total mass of A and B at any time remains constant and is equal to N. Hence, the mass of B, n, at time t can be expressed by the differential equation dn dt = k n (N – n). (d) The rate of increase of a population due to birth is αx and the rate of decrease due to death is βx2 where x is the population number at time t. The population number at any time can be found by solving the differential equation dx dt = αx – βx2 .
141 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 Example 14 The rate at which a substance evaporates is k times the amount of the substance that has not been evaporated. If the amount of the substance at the beginning is A and the amount that has been evaporated at time t is x, write down a differential equation involving x, A, k and t. (a) Solve this differential equation and sketch a graph of x against t. (b) Show that the time taken for half the amount to evaporate is 1 k ln 2. (c) Find the percentage of the amount of substance that has not been evaporated after time 1 2k ln 2. Solution: (a) The differential equation is dx dt = k(A – x). ∫ 1 (A – x) dx = ∫ k dt – ln | A – x | = kt + C When t = 0, x = 0, therefore C = –ln A. Hence, – ln | A – x | = kt – ln A In A – x A = –kt A – x A = e–kt x = A(1 – e–kt) (b) When half of the amount has evaporated, x = 1 2 A. Thus, A 2 = A(1 – e–kt) 1 – e–kt = 1 2 e–kt = 1 2 –kt = ln 1 2 t = 1 k ln 2 (c) When t = 1 2k ln 2, x = A(1 – e–—1 2 ln2 ) = A(1 – 2–—1 2 ) = 0.2929A Thus, the amount that has not evaporated is 0.7071A. Hence, 70.71% of the substance has not evaporated. x A t 0
142 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 Example 15 An object moves along a straight line and passes a fixed point O with velocity u in the positive direction of the x-axis. At time t, the object is at a displacement x from O and the velocity of the object is v. The rate of change of velocity has magnitude k v2 , where k is a constant, and is directed to towards the fixed point O. (a) Write down a differential equation for the motion of the object involving the velocity v and the time t. Hence, find the velocity v as a function of the time t. (b) Show that dv dt = v dv dx and hence, write down the differential equation for the motion of the object. Hence, find the velocity v as function of the displacement x. Hence, show that after a time t and the object has moved a distance x, 4kx = u4 – (u3 – 3kt) —4 3 . Solution: (a) The differential equation for the motion of the object involving the velocity v as a function of the time t is dv dt = – k v2 . (the negative sign shows that the rate of change of velocity is directed towards the fixed point O but the motion of the object is in the opposite direction i.e. positive direction of the x-axis) Separating the variables and integrating both sides: ∫ v2 dv = –∫ kdt v3 3 = –kt + C When t = 0, v = u, u3 3 = –k(0) + C ⇒ C = u3 3 Substituting the value of C v3 3 = –kt + u3 3 v3 = –3kt + u3 \ v = 3 –3kt + u3 (b) Using the chain rule of differentiation. dv dt = 1 dv dx 21 dx dt 2 The rate of change of x with respect to time t, dx dt = v \ dv dt = 1 dv dx 2v \ dv dt = v dv dx
143 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 The differential equation for the motion of the object that involves the change of velocity v with respect to the displacement x and the velocity v is v dv dx = – k v2 . Separating the variables and integrating both sides: ∫ v3 dv = –∫ k dx v4 4 = –kx + C When x = 0, v = u u4 4 = –k(0) + C ⇒ C = u4 4 Substituting the value of C, v4 4 = –kx + u4 4 v4 = –4kx + u4 \ v = 4 –4kx + u4 Equating the velocity found in (a) and that in (b), \ 4 –4kx + u4 = 3 –3kt + u3 –4kx + u4 = (u3 – 3kt) —4 3 \ 4kx = u4 – (u3 – 3kt) —4 3 Example 16 A research is being done on a particular species of bear on an animal reserve territory. Initially there are 50 bears on the reserve. After t years the number of bears, n, satisfies the differential equation dn dt = 1 20k n(k – n) where k is a constant. (a) Show that k = 200 if it is known that the rate of growth is 0.9 bear per year when n = 20. (b) What is the maximum rate of growth? (c) Explain what happens as n approaches 200? (d) Obtain the solution of the differential equation and sketch the solution curve. (e) Find (i) the number of bears after 77 years, (ii) the time for which the number of bears is 100.
144 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 Solution: (a) Substitute n = 20 and dn dt = 0.9 0.9 = 1 20k (20)(k – 20) 0.9k = k – 20 ⇒ k = 200 (b) \ The differential equation is dn dt = 1 4000 n(200 – n). The expression 1 4000 n(200 – n) is a quadratic expression and has the maximum value at n = 1 2 (0 + 200) = 100. The maximum value of 1 4000 n(200 – n) is 1 4000 (100)(200 – 100) = 2.5 Hence, the maximum growth rate is 2.5 bears per year. (c) As n approaches 200, the growth rate of the bears approaches 0, i.e. there will be no bear born. (d) Separating the variables and integrating both sides: ∫ 1 dn n(200 – n)2 = ∫ 1 4000 dt 1 200 ∫ 1 1 n + 1 (200 – n)2 dn = 1 4000 ∫ dt (lnn – ln(200 – n) = 0.05t + C ln1 n 200 – n 2 = 0.05t + C Substituting t = 0, n = 50 ln1 50 200 – 50 2 = 0.05(0) + C ⇒ C = ln 1 3 \ ln1 n 200 – n 2 = 0.05t + ln 1 3 ln1 n 200 – n 2 – ln 1 3 = 0.05t ln1 3n 200 – n 2 = 0.05t 1 3n 200 – n 2 = e0.05t 1 200 – n 3n 2 = e–0.05t 200 – n = 3ne–0.05t \ n = 200 1 + 3e–0.05t
145 Mathematics Semester 2 STPM Chapter 4 Differential Equations 4 The solution curve of n = 200 1 + 3e–0.05t . When t = 0, n = 50. As t → ∞, e–0.05t → 0, \ n → 200. n O t 200 50 (e) (i) Substitute t = 77 \ n = 200 1 + 3e–0.05 × 77 = 188 bears (ii) Substitute n = 100 and using the form ln1 3n 200 – n 2 = 0.05t t = 20 ln1 300 200 – 100 2 = 22 years Example 17 A series of regular interval injections is administered onto laboratory mice to test the efficacy of a new drug. The rate of destruction of the drug is proportional to the amount of drug present in the mice. (a) If k is the proportional constant, x is the amount of drug at time t, write a differential equation relating the amount of drug and the time and hence, show that the general solution is x = Ae–kt where A is an arbitrary constant. (b) Initially an amount, D, of the drug is injected onto a mice and after a time t = 1 hour the amount of the drug remaining is 3 4 D. Show that x = D1 3 4 2 t . (c) The drug is injected again onto the mice after t = 1 hour and t = 2 hours. Find the amount of drug remaining in the body immediately after 2 hours. (d) If the drug is administered at regular intervals of 1 hour for an indefinite period, find the amount of drug remaining in the mice. Solution: (a) The differential equation for amount of drug remaining after a time t is dx dt = –kx (The negative sign shows that the amount, x, is getting less with time, t) Separating and integrating both sides. ∫ 1 x dx = –∫ k dt lnx = –kt + C