# Properties of Definite Integrals Worksheet

Properties of Definite Integrals Worksheet
• Page 1
1.
Find the equivalent of ${\int }_{a}^{b}$ $f$($x$) $d$$x$ + ${\int }_{b}^{c}$ $f$($x$) $d$$x$.
 a. ${\int }_{b}^{a}$ $f$($x$) $d$$x$ b. ${\int }_{c}^{a}$ $f$($x$) $d$$x$ c. - ${\int }_{a}^{c}$ $f$($x$) $d$$x$ d. ${\int }_{a}^{c}$ $f$($x$) $d$$x$

#### Solution:

An integral depends additively on the interval of integration.

So, the area between a and b and the area between b and c equals the area between a and c.

Therefore, ab f(x) dx + bc f(x) dx = ac f(x) dx.

This illustrates the general property of continuous functions.

2.
Evaluate: ${\int }_{0}^{2}$ 5($x$ + 3) $d$$x$
 a. 8 b. 48 c. 13 d. 40

#### Solution:

02 5(x + 3) dx
[Given.]

ab kf(x) dx = kab f(x) dx
[Property of Definite Integral.]

02 5(x + 3) dx = 502 (x + 3) dx

= 5[x22 + 3x ]02

= 5[8]
[Apply the limits of integration.]

= 40
[Simplify.]

3.
Evaluate ${\int }_{3}^{2}$ $x$ + 5 $d$$x$ by interchanging the limits.
 a. - $\frac{15}{2}$ b. $\frac{13}{2}$ c. $\frac{15}{2}$ d. $\frac{5}{2}$

#### Solution:

32 (x + 5) dx
[Given.]

ba f(x) dx = - ab f(x) dx
[Interchanging the limits of integration changes the sign of the definite integral.]

32 (x + 5) dx = - 23 (x + 5) dx

= - [23 x dx + 23 5 dx]
[Use Sum rule of integration.]

= [- x22 + 5x ]23

= - 15 / 2
[Apply the limits of integration.]

4.
Evaluate: ${\int }_{-2}^{2}$ | $x$ | $\mathrm{dx}$
 a. 2 b. 8 c. 16 d. 4

#### Solution:

-22 | x | dx

= -20 | x | dx + 02| x | dx
[Use abf(x) dx = acf(x) dx + cbf(x) dx.]

= -20 - x dx + 02x dx
[| x | = x, if x > 0 and | x | = - x, if x < 0.]

=[- x22 ]-20 + [x22 ]02

= 2 + 2 = 4

So, -22 | x | dx = 4

5.
Evaluate: ${\int }_{0}^{\pi /4}$sin2 2$x$ $\mathrm{dx}$
 a. 2$\pi$ b. $\pi$ c. $\frac{\pi }{4}$ d. $\frac{\pi }{2}$ e. $\frac{\pi }{8}$

#### Solution:

Let I = 0π/4sin2 2x dx --- (1)

I = 0π/4sin22(π4-x) dx
[Use 0af(x) dx = 0af(a - x) dx.]

I = 0π/4sin2(π2 - 2x) dx

I = 0π/4cos22x dx --- (2)

I + I = 0π/4sin22x dx + 0π/4cos22x dx
[Add equation (1) and equation (2).]

2I = 0π/4(sin22x + cos22x) dx

2I = 0π/4 (1) dx = [ x ]0π/4
[Use sin22x + cos22x = 1.]

2I = π4 - 0 = π4

I = π8

So, 0π/4sin2 2x dx = π8

6.
Evaluate: ${\int }_{2}^{2}$$\frac{2x}{{x}^{2}+1}$ $\mathrm{dx}$
 a. 4 b. 2 ln 5 c. 2 d. ln 5

#### Solution:

222xx2+1 dx

= 0
[Use aaf(x) dx = 0.]

So, 222xx2+1 dx = 0

7.
Evaluate: ${\int }_{1}^{2}$($x$2 + $\frac{1}{{x}^{2}}$) $\mathrm{dx}$ and use it to determine the value of ${\int }_{2}^{1}$($x$2 + $\frac{1}{{x}^{2}}$) $\mathrm{dx}$
 a. $\frac{13}{6}$, - $\frac{13}{6}$ b. - $\frac{13}{6}$, $\frac{13}{6}$ c. $\frac{1}{2}$, - $\frac{1}{2}$ d. $\frac{5}{3}$, - $\frac{5}{3}$ e. $\frac{6}{13}$, - $\frac{6}{13}$

#### Solution:

12(x2 + 1x2) dx

= 12x2 dx + 121x2 dx
[Use Sum Rule of integration.]

= [ x33 ]12 + [- 1x ]12

= [8 / 3 - 1] + [- 1 / 2 + 1]

= 5 / 3 + 1 / 2 = 13 / 6

21(x2 + 1x2) dx = - 12(x2 + 1x2) dx
[Use baf(x) dx = - abf(x) dx.]

= - 13 / 6
[12(x2 + 1x2) dx = 13 / 6.]

So, 21(x2 + 1x2) dx = - 13 / 6

8.
Evaluate: ${\int }_{0}^{5}$ | $\frac{x}{2}$ - 1 | $\mathrm{dx}$
 a. $\frac{25}{4}$ b. 2 c. $\frac{17}{4}$ d. $\frac{5}{4}$ e. $\frac{13}{4}$

#### Solution:

05 | x2 - 1 | dx

= 02 | x2 - 1 | dx + 25 | x2 - 1 | dx
[Use ab f(x) dx = ac f(x) dx + cb f(x) dx]

= 02- (x2 - 1) dx + 25(x2 -1) dx
[( x2 - 1 ) is negative in [0, 2] and positive in [2, 5].]

= [- (x24 - x) ]02 + [ (x24 - x) ]25

= [- (1 - 2) + 0] + [ (25 / 4 - 5) - (1 - 2)]

= 1 + 5 / 4 + 1 = 13 / 4

So, 05 | x2 - 1 | dx = 13 / 4

9.
Evaluate: ${\int }_{0}^{1}$$x$$\sqrt{1-x}$ $\mathrm{dx}$
 a. $\frac{8}{15}$ b. $\frac{4}{15}$ c. 2 d. $\frac{2}{5}$ e. $\frac{2}{3}$

#### Solution:

01x1-x dx

= 01(1 - x)1-(1-x) dx
[Use 0af(x) dx = 0af(a - x) dx.]

= 01(1 - x)x dx

= 01x12 dx - 01x32 dx
[Use Difference Rule of integration.]

= [x3232 ]01 - [x5252 ]01

= 2 / 3 - 2 / 5
[Apply the limits of integration.]

= 4 / 15

So, 01x1-x dx = 4 / 15

10.
Evaluate: ${\int }_{0}^{\pi /2}$(1 + sin2$x$) $\mathrm{dx}$
 a. $\frac{3\pi }{4}$ b. $\frac{\pi }{4}$ c. $\frac{{3\pi }^{2}}{4}$ d. $\frac{3\pi }{2}$ e. $\frac{\pi }{2}$

#### Solution:

Let I = 0π/2(1 + sin2x) dx --- (1)

I = 0π/2(1 + sin2 (π2 - x)) dx
[Use 0af(x) dx = 0af(a - x) dx.]

I = 0π/2(1 + cos2 x) dx --- (2)

I + I = 0π/2(2 + cos2x + sin2x) dx
[Add equation (1) and equation (2).]

2I = 0π/23 dx
[Use cos2x + sin2x = 1.]

2I = 3 [ x ]0π/2

2I = 3 [π2 - 0]
[Apply the limits of integration.]

I = 3π4

So, 0π/2(1 + sin2x) dx = 3π4