﻿ Fundamental Theorem of Calculus Worksheet - Page 2 | Problems & Solutions

Fundamental Theorem of Calculus Worksheet - Page 2

Fundamental Theorem of Calculus Worksheet
• Page 2
11.
Use fundamental theorem of calculus, to evaluate ${\int }_{3}^{5}$$\frac{3}{{x}^{2}-x-2}$ $\mathrm{dx}$.
 a. 2 ln 2 b. 2 ln 3 + 3 ln 2 c. ln 2 d. ln 3 + 2 ln 2 e. ln 3 + ln 2

Solution:

3x2-x-2 = 3(x-2)(x+1)
[Factor the denominator.]

Let 3(x-2)(x+1) = Ax-2+Bx+1
[Use Partial fractions.]

3 = A(x + 1) + B(x - 2)

Put x = - 1, B = - 1

Put x = 2, A = 1

35 3x2-x-2 dx = 35(1x-2-1x+1) dx

= 351x-2 dx - 351x+1 dx
[Use Difference Rule of integration.]

= [ ln | x - 2 | ]35 - [ ln | x + 1 | ]35

= [ ln 3 - ln 1 ] - [ ln 6 - ln 4 ]
[Apply the limits of integration.]

= ln 3 - ln 6 + ln 4

= ln 3 - (ln 2 + ln 3) + 2 ln 2
[Use ln (ab) = ln a + ln b.]

= ln 2

So, 353x2-x-2 dx = ln 2

12.
Use fundamental theorem of calculus to evaluate ${\int }_{1}^{2}$$x$2 $e$2$x$ $\mathrm{dx}$.
 a. $\frac{{e}^{2}}{4}$ b. $\frac{{5e}^{2}-1}{4}$ c. $\frac{{5e}^{4}}{4}$ d. $\frac{{e}^{2}\left({5e}^{2}+1\right)}{4}$ e. $\frac{{e}^{2}\left({5e}^{2}-1\right)}{4}$

Solution:

12x2 e2x dx

= [x2e2x2 ]12 - 12 2xe2x2 dx
[Use integration by parts.]

= 2e4 - e22 - [xe2x2]12 + 12 e2x2 dx

= 2e4 - e22 - 1 / 2 [2e4 - e2] + [e2x4 ]12

= 2e4 - e22 - e4 + e22 + e44 - e24
[Apply the limits of integration.]

= 5e44 - e24

= 5e4-e24

= e2(5e2-1)4
[Simplify.]

13.
Use fundamental theorem of calculus, to evaluate ${\int }_{3}^{5}$$x$$\sqrt{{x}^{2}-9}$ $\mathrm{dx}$.
 a. 32 b. $\frac{64}{3}$ c. $\frac{128}{3}$ d. $\frac{16}{3}$ e. $\frac{4}{3}$

Solution:

Let x2 - 9 = u, 2x dx = du, x dx = 1 / 2 du

At x = 3, u = 0 and at x = 5, u = 16.

35xx2-9 dx = 1 / 2 016u du

= 1 / 2 [u3232 ]016

= 1 / 3 [ (16)32 - 0 ]
[Apply the limits of integration.]

= 64 / 3

14.
Use the fundamental theorem of calculus, to evaluate ${\int }_{0}^{4}$$\sqrt{16-{x}^{2}}$ $\mathrm{dx}$.
 a. 8 b. 4$\pi$ c. $\pi$ d. 4 e. $\frac{\pi }{2}$

Solution:

Let x = 4 sinθ, dx = 4 cosθ dθ

At x = 0, θ = 0 and at x = 4, θ = π / 2

0416-x2 dx = 0π/216-16sin2θ (4 cosθ) dθ

= 16 0π/2cos2θ dθ

= 8 0π/2(1 + cos 2θ) dθ
[Use 1 + cos 2θ = 2 cos2θ.]

= 8[θ + sin2θ2 ]0π/2

= 8 [π2] = 4π
[Apply the limits of integration.]

15.
Use the fundamental theorem of calculus, to evaluate ${\int }_{0}^{1}$$\frac{{x}^{2}}{1+{x}^{6}}$ $\mathrm{dx}$.
 a. $\frac{3\pi }{4}$ b. $\frac{\pi }{2}$ c. $\frac{\pi }{6}$ d. $\frac{\pi }{3}$ e. $\frac{\pi }{12}$

Solution:

Let x3 = t, 3x2 dx = dt, x2 dx = dt3

At x = 0, t = 0 and at x = 1, t = 1

01x21+x6 dx = 1 / 3 01dt1+t2

= 1 / 3 [tan-1(t) ]01

= 1 / 3 [tan -1(1) - tan -1(0)]
[Apply the limits of integration.]

= 1 / 3 (π4) = π12

16.
Use the fundamental theorem of calculus, to evaluate ${\int }_{0}^{1}$$\frac{{e}^{x}-{e}^{-x}}{2}$ $\mathrm{dx}$.
 a. $\frac{{\left(e+1\right)}^{2}}{2e}$ b. $\frac{{\left(e-1\right)}^{2}}{2e}$ c. $\frac{{\left(2e+1\right)}^{2}}{2e}$ d. $\frac{{\left(e-1\right)}^{2}}{2}$ e. $\frac{{\left(e-1\right)}^{2}}{e}$

Solution:

01ex-e-x2 dx

= 01ex2 dx - 01e-x2 dx
[Use Difference Rule of integration.]

= 1 / 2 [ex ]01 + 1 / 2 [e -x ]01

= 1 / 2 [e1 - eo] + 1 / 2 [e -1 - eo]
[Apply the limits of integration.]

= e2 - 1 / 2 + 12e - 1 / 2

= e2+12e - 1 = e2-2e+12e

= (e-1)22e

17.
Use the fundamental theorem of calculus, to evaluate ${\int }_{2}^{5}$$\frac{1}{{4x}^{2}-9}$ $\mathrm{dx}$.
 a. $\frac{1}{12}$ [ ln 7 + ln 13 ] b. $\frac{1}{12}$ (ln 13) c. $\frac{1}{6}$ (ln 7) d. $\frac{1}{12}$ [ 2 ln 7 + ln 13 ] e. $\frac{1}{12}$ [ 2 ln 7 - ln 13 ]

Solution:

2514x2-9 dx

= 251(2x-3)(2x+3) dx

= 25(162x-3-162x+3) dx
[Use the method of partial fractions.]

= 1 / 6 [ ln|2x-3|2-ln|2x+3|2 ]25

= 1 / 12 [ ln | 2x - 3 | - ln | 2x + 3 | ]25

= 1 / 12 [ ln 7 - ln 13 - ln 1 + ln 7 ]
[Apply the limits of integration.]

= 1 / 12 [2 ln 7 - ln 13 ]

18.
Use the fundamental theorem of calculus, to evaluate ${\int }_{1}^{e}$ln $x$ $\mathrm{dx}$.
 a. 2$e$ + 1 b. 2 c. $e$ d. 2$e$ e. 1

Solution:

1eln x dx

= [ x ln x - x ]1e
[Use integration by parts.]

= [ e ln e - e ] - [ ln 1 - 1 ]
[Apply the limits of integration.]

= 1
[Simplify.]

19.
Use the fundamental theorem of calculus, to evaluate ${\int }_{0}^{\pi }$sin ($\frac{x}{2}$) $\mathrm{dx}$.
 a. 2 b. 1 c. 4 d. - 2

Solution:

Let x2 = t, dx = 2 dt

At x = 0, t = 0 and at x = π, t = π2

0πsin (x2) dx = 2 0π/2sin t dt

= 2 [ - cos t ]0π/2

= 2 [- cos π2 + cos 0 ]
[Apply the limits of integration.]

= 2 (1) = 2

20.
Use the fundamental theorem of calculus, to evaluate ${\int }_{0}^{\pi /2}$sin 4$x$ cos $x$ $\mathrm{dx}$.
 a. $\frac{4}{15}$ b. 8 c. $\frac{8}{15}$ d. $\frac{1}{6}$ e. $\frac{1}{15}$

Solution:

0π/2sin 4x cos x dx

= 1 / 2 0π/2(sin 5x + sin 3x) dx
[Use sin(A + B) + sin(A - B) = 2 sin A cos B.]

= 1 / 2 [0π/2sin 5x dx + 0π/2sin 3x dx]
[Use Sum Rule of integration.]

= 1 / 2 [[-cos5x5 ]0π/2 + [-cos3x3 ]0π/2]

= 1 / 2 [ 1 / 5 (- cos5π2 + cos 0) + 1 / 3 (- cos3π2 + cos 0)]
[Apply the limits of integration.]

= 1 / 2 [1 / 5 (0 + 1) + 1 / 3 (1)]

= 1 / 2 [1 / 5 + 1 / 3 ]

= 1 / 2 [815]

= 4 / 15