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Fundamental Theorem of Calculus Worksheet - Page 2

Fundamental Theorem of Calculus Worksheet
  • Page 2
 11.  
Use fundamental theorem of calculus, to evaluate 353x2-x-2 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


Correct answer : (3)
 12.  
Use fundamental theorem of calculus to evaluate 12x2 e2x dx.
a.
e24
b.
5e2-14
c.
5e44
d.
e2(5e2+1)4
e.
e2(5e2-1)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.]


Correct answer : (5)
 13.  
Use fundamental theorem of calculus, to evaluate 35xx2-9 dx.
a.
32
b.
64 3
c.
128 3
d.
16 3
e.
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


Correct answer : (2)
 14.  
Use the fundamental theorem of calculus, to evaluate 0416-x2 dx.
a.
8
b.
4π
c.
π
d.
4
e.
π 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.]


Correct answer : (2)
 15.  
Use the fundamental theorem of calculus, to evaluate 01x21+x6 dx.
a.
3π4
b.
π2
c.
π6
d.
π3
e.
π12


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


Correct answer : (5)
 16.  
Use the fundamental theorem of calculus, to evaluate 01ex-e-x2 dx.
a.
(e+1)22e
b.
(e-1)22e
c.
(2e+1)22e
d.
(e-1)22
e.
(e-1)2e


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


Correct answer : (2)
 17.  
Use the fundamental theorem of calculus, to evaluate 2514x2-9 dx.
a.
1 12 [ ln 7 + ln 13 ]
b.
1 12 (ln 13)
c.
1 6 (ln 7)
d.
1 12 [ 2 ln 7 + ln 13 ]
e.
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 ]


Correct answer : (5)
 18.  
Use the fundamental theorem of calculus, to evaluate 1eln x dx.
a.
2e + 1
b.
2
c.
e
d.
2e
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.]


Correct answer : (5)
 19.  
Use the fundamental theorem of calculus, to evaluate 0πsin (x2) 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


Correct answer : (1)
 20.  
Use the fundamental theorem of calculus, to evaluate 0π/2sin 4x cos x dx.
a.
4 15
b.
8
c.
8 15
d.
1 6
e.
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


Correct answer : (1)

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