﻿ Fundamental Theorem of Calculus Worksheet - Page 3 | Problems & Solutions

# Fundamental Theorem of Calculus Worksheet - Page 3

Fundamental Theorem of Calculus Worksheet
• Page 3
21.
Use the fundamental theorem of calculus, to evaluate ${\int }_{2}^{4}$ $\frac{{4x}^{2}-1}{{2x}^{2}+3x-2}$ $\mathrm{dx}$.
 a. 4 + ln 2 b. 4 c. 4 - ln 2 d. 4 - 3 ln $\frac{3}{2}$ e. 4 + 3 ln $\frac{3}{2}$

#### Solution:

24 4x2-12x2+3x-2 dx

= 24 (2x-1)(2x+1)(2x-1)(x+2) dx
[Factor both numerator and denominator.]

= 242x+1x+2 dx

= 24 2x+4-3x+2 dx
[Add and subtract 3 in the numerator.]

= 24 2(x+2)(x+2) dx - 3 241x+2 dx
[Use Difference Rule of integration.]

= 2 [ x ]24 - 3 [ ln | x + 2 | ]24

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

= 4 - 3 ln 32

22.
Use the fundamental theorem of calculus, to evaluate ${\int }_{0}^{ln 2}$$\frac{{e}^{4x}-1}{{e}^{4x}+1}$ $\mathrm{dx}$.
 a. $\frac{1}{2}$ ln ($\frac{1}{8}$) b. $\frac{1}{2}$ ln (2) c. $\frac{1}{2}$ln ($\frac{17}{8}$) d. ln ($\frac{17}{8}$) e. ln ($\frac{1}{8}$)

#### Solution:

0ln 2e4x-1e4x+1 dx

= 0ln 2e4x+1-2e4x+1 dx
[Rearrange the numerator.]

= 0ln 2e4x+1e4x+1 dx - 2 0ln 21e4x+1 dx
[Use Difference Rule of integration.]

= 0ln 2dx - 2 0ln 21e4x+1 dx

= ln 2 - 2 0ln 2 e-4x1+e-4x dx
[Multiply both numerator and denominator by e- 4x.]

= ln 2 - 2[ - 1 / 4 ln (1 + e- 4x) ]0ln 2
[Evaluate the integral.]

= ln 2 + 1 / 2 (ln 17 / 16 - ln 2)

= 1 / 2 (ln 2 + ln 17 / 16 ) = 1 / 2ln (17 / 8)
[Use ln a + ln b = ln (ab).]

23.
Use the fundamental theorem of calculus, to evaluate ${\int }_{0}^{\pi /2}$$e$2$x$ sin 2$x$ $\mathrm{dx}$.
 a. $\frac{{e}^{\pi }}{2}$ b. ${e}^{\pi }$ c. ${e}^{\frac{\pi }{2}}$ d. 2${e}^{\pi }$ e. $\frac{{e}^{\pi }}{4}$

#### Solution:

Let I = e2x sin 2x dx

= sin 2x (e2x2) - (e2x2)(cos 2x) (2) dx
[Use integration by parts.]

= e2xsin2x2 - [e2xcos2x2 + e2xsin 2x dx]
[Use integration by parts again.]

= e2xsin2x2-e2xcos2x2 - I

2I = e2x2 [sin 2x - cos 2x]

I = e2x4 [sin 2x - cos 2x]

0π/2e2xsin 2x dx = [e2x4 [sin 2x - cos 2x] ]0π/2

= eπ4 [(sinπ - cosπ) - (sin 0 - cos 0)]
[Apply the limits of integration.]

= eπ4 [ 1 + 1 ]

= eπ2

24.
Use the fundamental theorem of calculus to evaluate ${\int }_{0}^{2}$$\frac{1}{1+{2}^{x}}$ $\mathrm{dx}$.
 a. 1 - $\frac{\mathrm{ln}5}{\mathrm{ln}2}$ b. 2 ln 2 - ln 5 c. 3 - $\frac{\mathrm{ln}5}{\mathrm{ln}2}$ d. 3 + $\frac{\mathrm{ln}5}{\mathrm{ln}2}$ e. 3 ln 2 - ln 5

#### Solution:

Let 1 + 2x = t, 2x ln 2 dx = dt, so dx = dt(t-1)ln 2

At x = 0, t = 2 and at x = 2, t = 5

02 11+2x dx = 25dtt(t-1)ln 2

= 1ln 2 251t(t-1) dt

= 1ln 2 25(1t-1-1t) dt
[Use the method of partial fractions.]

= 1ln 2[(ln | t - 1 | )25 - (ln | t | )25]

= 1ln 2 [ln 4 - ln 1 - ln 5 + ln 2]
[Apply the limits of integration.]

= 1ln 2 [ 2 ln 2 - ln 5 + ln 2 ]

= 1ln 2 [ 3 ln 2 - ln 5 ]

= 3 - ln 5ln 2

25.
Use the fundamental theorem of calculus, to evaluate ${\int }_{1}^{2}$$\frac{3}{x\left({x}^{3}+8\right)}$ $\mathrm{dx}$.
 a. $\frac{1}{4}$ (ln 3 + ln 2) b. $\frac{1}{4}$ ln $\sqrt{2}$ c. $\frac{1}{4}$ (ln 3 - $\frac{1}{2}$ ln 2) d. $\frac{1}{4}$ ln 3 e. $\frac{1}{4}$ (ln 3 + $\frac{1}{2}$ ln 2)

#### Solution:

123x(x3+8) dx

= 123 / 8 [1x-x2x3+8] dx
[Use the method of partial fractions.]

= 3 / 8 121x dx - 1 / 8 123x2x3+8 dx
[Use Difference Rule of integration.]

= 3 / 8 [ln | x | ]12 - 1 / 8 [ln | x3 + 8 | ]12

= 3 / 8 [ ln 2 - ln 1 ] - 1 / 8 [ ln 16 - ln 9 ]
[Apply the limits of integration.]

= 3 / 8 ln 2 - 1 / 2 ln 2 + 2 / 8 ln 3

= - 1 / 8 ln 2 + 1 / 4 ln 3

= 1 / 4 (ln 3 - 1 / 2 ln 2)