﻿ The Fundamental Theorem of Calculus Worksheet | Problems & Solutions

# The Fundamental Theorem of Calculus Worksheet

The Fundamental Theorem of Calculus Worksheet
• Page 1
1.
Using the fundamental theorem of calculus, evaluate ${\int }_{o}^{\pi }$sin 7$x$ $d$$x$.
 a. - $\frac{2}{7}$ b. $\frac{2}{7}$ c. 2

#### Solution:

o πsin 7x dx = 1 / 7 (- cos 7x)|0π

= 1 / 7 [(- cos 7π) - (- cos 0)]
[Evaluate the limits.]

= 1 / 7 (1 + 1) = 2 / 7

2.
Using the fundamental theorem of calculus, evaluate ${\int }_{0}^{\pi }$(7$e$$x$ - 4cos $x$) $d$$x$.
 a. 7$e$π b. 7[$e$π - 1] c. 7$e$π - 1 d. 7[$e$π + 1]

#### Solution:

0π(7ex - 4cos x) dx = 70π(ex) dx + 0π(- 4cos x) dx

= 70π(ex) dx - 40π(cos x) dx

= 7(ex) |0π - 4(sin x) |0π

= 7[eπ - e0] - 4 [sin π - sin 0]
[Evaluate the limits.]

= 7[eπ - 1]

3.
Using the fundamental theorem of calculus, evaluate ${\int }_{0}^{10}$|$x$2 - 5$x$| $d$$x$.
 a. 300 b. 500 c. 125 d. $\frac{250}{3}$

#### Solution:

The given function can be rewritten as,

|x2 - 5x| = 5x - x2 if 0 ≤ x ≤ 5 and = x2 - 5x if 5 ≤ x ≤ 10

Hence, 010|x2 - 5x| dx = 05(5x - x2) dx + 510(x2 - 5x) dx

= 5 05x dx - 05x2 dx + 510x2 dx - 5 510x dx

= 5 x22 |05 - x33 |05 + x33 |510 - 5 x22 |510

= 5 / 2[25 - 0] - 1 / 3[125 - 0] + 1 / 3[1000 - 125] - 5 / 2[100 - 25]
[Evaluate the limits.]

= 1252 -1253 +8753 -3752

= 250 - 125 = 125.

4.
Evaluate ${\int }_{0}^{2}$(8$x$3 - 6$x$) $d$$x$.
 a. 104 b. - 20 c. 20 d. 90

#### Solution:

02(8x3 - 6x) dx = [8 (x44 )- 6 (x22 )]02

= [32 - 12 ] - [0 - 0]
[Evaluate the limits.]

= 20

5.
Evaluate ${\int }_{5}^{6}$(8$x$2 + 8$x$-1 + 6sin $x$) $\mathrm{dx}$.
 a. 14.32 b. 1.45 c. 244.11 d. 12.87

#### Solution:

56(8x2 + 8x-1 + 6sin x) dx = 856x2 dx + 856x-1 dx + 656sin x dx

= 8.x33 |56 + 8.ln x |56 + 6(-cos x)|56

= 8 [633 -533] + 8[ln 6 - ln 5] - 6cos 6 + 6cos 5

= 728 / 3+ 14.32 - 12.87 - 6 + 6
[Evaluate the limits.]

244.11
[Simplify.]

6.
Using the fundamental theorem of calculus, evaluate ${\int }_{0.5}^{0.6}$(3$x$-1 + 4cos $x$ + 6$\frac{1}{\sqrt[]{{\mathrm{1 - x}}_{}^{2}}}$ ) $d$$x$.
 a. 1.607 b. 1.067 c. 1.76 d. 1.706

#### Solution:

0.50.6(3x-1 + 4cos x + 6121 - x2 )dx = 0.50.63x-1 dx + 40.50.64cos x dx + 60.50.611 - x2 dx

= 3ln x |0.50.6 + 4sin x |0.50.6 + 6sin-1 x |0.50.6

= 3[ln (0.6) - ln (0.5)] + 4[sin (0.6) - sin (0.5)] + 6[sin-1 (0.6) - sin-1 (0.5)]
[Evaluate the limits.]

1.607
[Simplify.]

7.
Using the fundamental theorem of calculus, evaluate ${\int }_{-0.5}^{0.1}$(3 + .
 a. 3 b. -1 c. -2 d. 2

#### Solution:

-0.50.1(3 + 11 -x2 + 2cos x ) dx = 3 -0.50.11 dx + -0.50.1 11 -x2dx + 2 -0.50.1 cos x dx

= 3 x |-0.50.1 + sin-1x |-0.50.1 + 2 sin x |-0.50.1

= 3[0.1 + 0.5] + sin-1 (0.1) - sin-1(-0.5) + 2[sin(0.1) - sin(-0.5)]
[Evaluate the limits.]

3
[Simplify.]

8.
Using the fundamental theorem of calculus, evaluate ${\int }_{-2}^{2}$(15$x$2 - 8$x$ + 9) $d$$x$.
 a. 204 b. -23 c. -120 d. 116

#### Solution:

-22(15x2 - 8x + 9) dx = (5x3 - 4x2 + 9x) |-22

= [5(2)3 - 4(2)2 + 9(2)] - [5(-2)3 - 4(-2)2 + 9(-2)]

= (40 - 16 + 18) - (- 40 - 16 - 18)
[Evaluate the limits.]

= 42 - (- 74)

= 116

9.
Using the fundamental theorem of calculus, evaluate ${\int }_{0}^{3}$5 $\mathrm{du}$.
 a. $\frac{65}{2}$ b. $\frac{135}{4}$ c. $\frac{5}{4}$ d. 35

#### Solution:

035 83 u+1 du
[Given.]

Let w = 8 / 3u + 1. Then dw = 8 / 3du. If u = 3, w = 9. If u = 0, w = 1.

So, 035 83u + 1 du = 15 / 8 03(83 u + 1)12 8 / 3du

= 15 / 819w1/2 dw

= 15 / 8 (2 / 3) (w)3/2 |19

= 5 / 4 (9)3/2 - 5 / 4(1)3/2
[Evaluate the limits.]

= 135 / 4 - 5 / 4 = 65 / 2

10.
Using the fundamental theorem of calculus, evaluate ${\int }_{0}^{1}$2($t$ $\frac{1}{2}$ - $t$) $\mathrm{dt}$.
 a. -3 b. 3 c. - $\frac{1}{3}$ d. $\frac{1}{3}$

#### Solution:

012(t 1 / 2 - t) dt = 2[t3/2 3/2 - t2 2 ] |01

= 2[2 / 3t3/2 - 1 / 2t2] |01

= 2[2 / 3(1)3/2 - 1 / 2(1)2] - 2[2 / 3(0)3/2 - 1 / 2(0)2]
[Evaluate the limits.]

= 2(2 / 3 - 1 / 2) - 2(0 - 0)

= 2(1 / 6) = 1 / 3