4.6 Exercises 4.6.1 The Indefinite Integral Exercise 4.1 (mathematica/calculus/integral/fundamental.nb) Evaluate  (2x + 3) 10 dx. Hint, Solution Exercise 4.2 (mathematica/calculus/integral/fundamental.nb) Evaluate  (ln x) 2 x dx. Hint, Solution Exercise 4.3 (mathematica/calculus/integral/fundamental.nb) Evaluate  x √ x 2 + 3 dx. Hint, Solution Exercise 4.4 (mathematica/calculus/integral/fundamental.nb) Evaluate  cos x sin x dx. Hint, Solution Exercise 4.5 (mathematica/calculus/integral/fundamental.nb) Evaluate  x 2 x 3 −5 dx. Hint, Solution 4.6.2 The Definite Integral Exercise 4.6 (mathematica/calculus/integral/definite.nb) Use the result  b a f(x) dx = lim N→∞ N−1  n=0 f(x n )∆x 134 where ∆x = b−a N and x n = a + n∆x, to show that  1 0 x dx = 1 2 . Hint, Solution Exercise 4.7 (mathematica/calculus/integral/definite.nb) Evaluate the following integral using integration by parts and the Pythagorean identity.  π 0 sin 2 x dx Hint, Solution Exercise 4.8 (mathematica/calculus/integral/definite.nb) Prove that d dx  f(x) g(x) h(ξ) dξ = h(f(x))f  (x) − h(g(x))g  (x). (Don’t use the limit definition of differentiation, use the Fundamental Theorem of Integral Calculus.) Hint, Solution Exercise 4.9 (mathematica/calculus/integral/definite.nb) Let A n be the area between the curves x and x n on the interval [0 . . . 1]. What is lim n→∞ A n ? Explai n this result geometrically. Hint, Solution Exercise 4.10 (mathematica/calculus/integral/taylor.nb) a. Show that f(x) = f(0) +  x 0 f  (x − ξ) dξ. b. From the above identity show that f(x) = f(0) + xf  (0) +  x 0 ξf  (x − ξ) dξ. 135 c. Using induction, show that f(x) = f(0) + xf  (0) + 1 2 x 2 f  (0) + ··· + 1 n! x n f (n) (0) +  x 0 1 n! ξ n f (n+1) (x − ξ) dξ. Hint, Solution Exercise 4.11 Find a function f(x) whose arc length from 0 to x is 2x. Hint, Solution Exercise 4.12 Consider a curve C, bounded by −1 and 1, on the interval (−1 . . . 1). Can the length of C be unbounded? What if we change to the closed interval [−1 . . . 1]? Hint, Solution 4.6.3 The Fundamental Theorem of Integration 4.6.4 Techniques of Integration Exercise 4.13 (mathematica/calculus/integral/parts.nb) Evaluate  x sin x dx. Hint, Solution Exercise 4.14 (mathematica/calculus/integral/parts.nb) Evaluate  x 3 e 2x dx. Hint, Solution Exercise 4.15 (mathematica/calculus/integral/partial.nb) Evaluate  1 x 2 −4 dx. Hint, Solution 136 Exercise 4.16 (mathematica/calculus/integral/partial.nb) Evaluate  x+1 x 3 +x 2 −6x dx. Hint, Solution 4.6.5 Improper Integrals Exercise 4.17 (mathematica/calculus/integral/improper.nb) Evaluate  4 0 1 (x−1) 2 dx. Hint, Solution Exercise 4.18 (mathematica/calculus/integral/improper.nb) Evaluate  1 0 1 √ x dx. Hint, Solution Exercise 4.19 (mathematica/calculus/integral/improper.nb) Evaluate  ∞ 0 1 x 2 +4 dx. Hint, Solution 137 4.7 Hints Hint 4.1 Make the change of variables u = 2x + 3. Hint 4.2 Make the change of variables u = ln x. Hint 4.3 Make the change of variables u = x 2 + 3. Hint 4.4 Make the change of variables u = sin x. Hint 4.5 Make the change of variables u = x 3 − 5. Hint 4.6  1 0 x dx = lim N→∞ N−1  n=0 x n ∆x = lim N→∞ N−1  n=0 (n∆x)∆x Hint 4.7 Let u = sin x and dv = sin x dx. Integration by parts will give you an equation for  π 0 sin 2 x dx. Hint 4.8 Let H  (x) = h(x) and evaluate the integral in terms of H(x). 138 Hint 4.9 CONTINUE Hint 4.10 a. Evaluate the integral. b. Use integration by parts to evaluate the integral. c. Use integration by parts with u = f (n+1) (x − ξ) and dv = 1 n! ξ n . Hint 4.11 The arc length from 0 to x is  x 0  1 + (f  (ξ)) 2 dξ (4.3) First show that the arc length of f(x) from a to b is 2(b −a). Then conclude that the integrand in Equation 4.3 must everywhere be 2. Hint 4.12 CONTINUE Hint 4.13 Let u = x, and dv = sin x dx. Hint 4.14 Perform integration by parts three succes sive times. For the first one let u = x 3 and dv = e 2x dx. Hint 4.15 Expanding the integrand in partial fractions, 1 x 2 − 4 = 1 (x − 2)(x + 2) = a (x − 2) + b (x + 2) 1 = a(x + 2) + b(x −2) 139 Set x = 2 and x = −2 to solve for a and b. Hint 4.16 Expanding the integral in partial fractions, x + 1 x 3 + x 2 − 6x = x + 1 x(x − 2)(x + 3) = a x + b x − 2 + c x + 3 x + 1 = a(x −2)(x + 3) + bx(x + 3) + cx(x −2) Set x = 0, x = 2 and x = −3 to solve for a, b and c. Hint 4.17  4 0 1 (x − 1) 2 dx = lim δ→0 +  1−δ 0 1 (x − 1) 2 dx + lim →0 +  4 1+ 1 (x − 1) 2 dx Hint 4.18  1 0 1 √ x dx = lim →0 +  1  1 √ x dx Hint 4.19  1 x 2 + a 2 dx = 1 a arctan  x a  140 4.8 Solutions Solution 4.1  (2x + 3) 10 dx Let u = 2x + 3, g(u) = x = u−3 2 , g  (u) = 1 2 .  (2x + 3) 10 dx =  u 10 1 2 du = u 11 11 1 2 = (2x + 3) 11 22 Solution 4.2  (ln x) 2 x dx =  (ln x) 2 d(ln x) dx dx = (ln x) 3 3 Solution 4.3  x √ x 2 + 3 dx =  √ x 2 + 3 1 2 d(x 2 ) dx dx = 1 2 (x 2 + 3) 3/2 3/2 = (x 2 + 3) 3/2 3 141 Solution 4.4  cos x sin x dx =  1 sin x d(sin x) dx dx = ln |sin x| Solution 4.5  x 2 x 3 − 5 dx =  1 x 3 − 5 1 3 d(x 3 ) dx dx = 1 3 ln |x 3 − 5| Solution 4.6  1 0 x dx = lim N→∞ N−1  n=0 x n ∆x = lim N→∞ N−1  n=0 (n∆x)∆x = lim N→∞ ∆x 2 N−1  n=0 n = lim N→∞ ∆x 2 N(N − 1) 2 = lim N→∞ N(N − 1) 2N 2 = 1 2 142 [...]... (x2 ) x 15 1 Solution 4.4 First we expand the integrand in partial fractions a b c 1 + x + x2 = + + (x + 1) 3 (x + 1) 3 (x + 1) 2 x + 1 a = (1 + x + x2 ) b= c= x= 1 =1 1 1! d (1 + x + x2 ) dx 1 2! d2 (1 + x + x2 ) dx2 = (1 + 2x) x= 1 = 1 x= 1 = x= 1 1 (2) 2 x= 1 =1 Then we can do the integration 1 + x + x2 dx = (x + 1) 3 1 1 1 − + 3 2 (x + 1) (x + 1) x +1 1 1 + + ln |x + 1| =− 2(x + 1) 2 x + 1 x + 1/ 2 + ln... + 3) 3 2 1 ln |x − 2| − ln |x + 3| + C = − ln |x| + 6 10 15 |x − 2|3 /10 = ln 1/ 6 +C |x| |x + 3|2 / 15 − Solution 4 .17 4 0 1 dx = lim δ→0+ (x − 1) 2 1 δ 0 1 dx + lim →0+ (x − 1) 2 1+ 1 δ→0 →0 x 1 0 1 1 1 = lim − 1 + lim − + + + δ→0 →0 δ 3 =∞+∞ = lim − + 1 x 1 1−δ 4 The integral diverges 14 8 + lim − + 1 dx (x − 1) 2 4 1+ Solution 4 .18 1 0 1 1 √ dx = lim →0+ x 1 √ dx x √ 1 = lim 2 x →0+ √ = lim 2 (1 − ) + →0... ln |x + 1| = (x + 1) 2 dx Solution 4 .5 Let f (x) be continuous Then b f (x) dx = (b − a)f (ξ), a for some ξ ∈ [a b] Solution 4.6 1 a b c d = + + + x(x − 1) (x − 2)(x − 3) x x 1 x−2 x−3 15 2 1 1 =− (0 − 1) (0 − 2)(0 − 3) 6 1 1 = b= (1) (1 − 2) (1 − 3) 2 1 1 c= =− (2)(2 − 1) (2 − 3) 2 1 1 = d= (3)(3 − 1) (3 − 2) 6 a= 1 1 1 1 1 =− + − + x(x − 1) (x − 2)(x − 3) 6x 2(x − 1) 2(x − 2) 6(x − 3) 15 3 Chapter 5 Vector... it holds for some n = m ≥ 0 x 1 1 1 n (n +1) f (x) = f (0) + xf (0) + x2 f (0) + · · · + xn f (n) (0) + ξ f (x − ξ) dξ 2 n! 0 n! 1 1 1 ξ n +1 f (n +1) (x − ξ) = f (0) + xf (0) + x2 f (0) + · · · + xn f (n) (0) + 2 n! (n + 1) ! x 1 − − ξ n +1 f (n+2) (x − ξ) dξ (n + 1) ! 0 1 1 1 = f (0) + xf (0) + x2 f (0) + · · · + xn f (n) (0) + xn +1 f (n +1) (0) 2 n! (n + 1) ! x 1 + ξ n +1 f (n+2) (x − ξ) dξ (n + 1) ! 0 x 0... 1 1 − dx 4(x − 2) 4(x + 2) 1 1 = ln |x − 2| − ln |x + 2| + C 4 4 1 x−2 = +C 4 x+2 14 7 Solution 4 .16 Expanding the integral in partial fractions, x3 x +1 x +1 A B C = = + + 2 − 6x +x x(x − 2)(x + 3) x x−2 x+3 x + 1 = A(x − 2)(x + 3) + Bx(x + 3) + Cx(x − 2) 1 Setting x = 0 yields A = − 6 Setting x = 2 yields B = x3 x +1 dx = + x2 − 6x 3 10 2 Setting x = −3 yields C = − 15 1 3 2 + − dx 6x 10 (x − 2) 15 (x... integer n 1 x2 xn +1 (x − x ) dx = − 2 n +1 1 n An = 0 14 3 = 0 1 1 − 2 n +1 Then we consider the area in the limit as n → ∞ 1 1 − 2 n +1 lim An = lim n→∞ n→∞ = 1 2 In Figure 4.3 we plot the functions x1 , x2 , x4 , x8 , , x1024 In the limit as n → ∞, xn on the interval [0 1] tends to the function 0 0≤x . solve for a, b and c. Hint 4 .17  4 0 1 (x − 1) 2 dx = lim δ→0 +  1 δ 0 1 (x − 1) 2 dx + lim →0 +  4 1+  1 (x − 1) 2 dx Hint 4 .18  1 0 1 √ x dx = lim →0 +  1  1 √ x dx Hint 4 .19  1 x 2 +. 2| 3 /10 |x| 1/ 6 |x + 3| 2 / 15 + C Solution 4 .17  4 0 1 (x − 1) 2 dx = lim δ→0 +  1 δ 0 1 (x − 1) 2 dx + lim →0 +  4 1+  1 (x − 1) 2 dx = lim δ→0 +  − 1 x − 1  1 δ 0 + lim →0 +  − 1 x − 1  4 1+  =. integrand in partial fractions. 1 + x + x 2 (x + 1) 3 = a (x + 1) 3 + b (x + 1) 2 + c x + 1 a = (1 + x + x 2 )   x= 1 = 1 b = 1 1!  d dx (1 + x + x 2 )      x= 1 = (1 + 2x)   x= 1 = 1 c