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Properties of Definite Integrals Worksheet

Properties of Definite Integrals Worksheet
  • Page 1
 1.  
Find the equivalent of ab f(x) dx + bc f(x) dx.
a.
ba f(x) dx
b.
ca f(x) dx
c.
- ac f(x) dx
d.
ac f(x) dx


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.


Correct answer : (4)
 2.  
Evaluate: 02 5(x + 3) dx
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.]


Correct answer : (4)
 3.  
Evaluate 32 x + 5 dx by interchanging the limits.
a.
- 15 2
b.
13 2
c.
15 2
d.
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.]


Correct answer : (1)
 4.  
Evaluate: -22 | x | 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


Correct answer : (5)
 5.  
Evaluate: 0π/4sin2 2x dx
a.
2π
b.
π
c.
π4
d.
π2
e.
π8


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


Correct answer : (5)
 6.  
Evaluate: 222xx2+1 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


Correct answer : (4)
 7.  
Evaluate: 12(x2 + 1x2) dx and use it to determine the value of 21(x2 + 1x2) dx
a.
13 6, - 13 6
b.
- 13 6, 13 6
c.
1 2, - 1 2
d.
5 3, - 5 3
e.
6 13, - 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


Correct answer : (1)
 8.  
Evaluate: 05 | x2 - 1 | dx
a.
25 4
b.
2
c.
17 4
d.
5 4
e.
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


Correct answer : (5)
 9.  
Evaluate: 01x1-x dx
a.
8 15
b.
4 15
c.
2
d.
2 5
e.
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


Correct answer : (2)
 10.  
Evaluate: 0π/2(1 + sin2x) dx
a.
3π4
b.
π4
c.
3π24
d.
3π2
e.
π2


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


Correct answer : (1)

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