Calculus 221 worksheet
Antiderivatives Solution
Exercise
Z
1
3ex dx.
1. Evaluate the integral
0
Solution:
1
Z
1
3e dx = (3e ) x
0
x
0
1
= 3e − 3e0 = 3e − 3.
2.
1
Z
Evaluate the integral
0
1
√
dx.
1 − x2
Solution:
Z
0
Z
3. Evaluate the integral
Solution:
Z
x
2013
+
√
2013
1
1
1
√
dx = (arcsin x) 1 − x2
0
= arcsin 1 − arcsin 0
π
= .
2
x2013 +
√
2013
1
√ dx =
x + 2013
x
x+
Z
1
√ dx.
2013
x
x2013 + x1/2013 + x−1/2013 dx
x2014
x2014/2013
x2012/2013
+
+
+C
2014 2014/2013 2012/2013
x2014 2013x2014/2013 2013x2012/2013
=
+
+
+ C.
2014
2014
2012
=
Z
4. Evaluate the integral
π
(ex + sin x) dx.
0
Solution:
Z
0
π
π
(e + sin x) dx = (e − cos x) x
x
0
π
= (e − cos π) − (e0 − cos 0)
= eπ + 1.
1
2
Z
1
5. Evaluate the integral
x dx and draw a sketch of the area the integral represents.
−1
Solution:
Z
1
x dx =
−1
x2
2
1
−1
2
1
(−1)2
−
2
2
= 0.
=
The integral represents the area:
Z
6. Evaluate the integral
1
ex dx and draw a sketch of the area the integral represents.
0
Solution:
Z
0
1
1
ex dx = (ex ) 0
= e1 − e0
= e − 1.
The integral represents the area:
3
Z
π
sin(x) dx and draw a sketch of the area the integral represents.
7. Evaluate the integral
0
Solution:
π
Z
0
π
sin(x) dx = (− cos(x)) 0
= − cos(π) + cos(0)
= 2.
The integral represents the area:
Z
2π
sin(x) dx and draw a sketch of the area the integral represents.
8. Evaluate the integral
0
Solution:
Z
2π
0
2π
sin(x) dx = (− cos(x)) 0
= − cos(2π) + cos(0)
= 0.
The integral represents the area:
4
9. Determine the area between y = xn and the x-axis, between x = 0 and x = 1.
Z
Solution: The region described is represented by the integral
1
xn dx.
0
n+1 1
Z 1
x
xn dx =
n+1 0
0
1n+1
0n+1
=
−
n+1 n+1
1
=
.
n+1
The area under the curve is
1
.
n+1
10. Determine the area under one arch (i.e., a half-period) of y = sin(6x).
Solution: One arch of y = sin(6x) runs from x = 0 to the next time sin(6x) = 0, which is
Z π/6
π
when 6x = π, or x = 6 . The region described is represented by the integral
sin(6x) dx.
0
π/6
1
sin(6x) dx = − cos(6x) 6
0
0
1
1
= − cos(6(π/6)) + cos(0)
6
6
1
= .
3
1
The area under the curve is .
3
Z
π/6
11. Determine the area above the x-axis and below the graph of y = 4x2 − x4 .
5
Solution: The curve y = 4x2 − x4 intersects the x-axis at x
Z = −2, 0, 2. The region described
2
4x2 − x4 dx.
is between −2 and 2, so it is represented by the integral
−2
Z
2
4x2 − x4 dx =
−2
2
4 3 1 5 x − x 3
5
−2
4
1
4
1
= (2)3 − (2)5 − (−2)3 + (−2)5
3
5
3
5
128
=
.
15
The area under the curve is
128
.
15
12. Determine the area underneath y = 1/x between x = a and x = b (where 0 < a < b).
Z b
1
dx.
Solution: The region described is represented by the integral
a x
b
Z b
1
dx = (ln |x|) a x
a
= ln(b) − ln(a)
b
= ln
.
a
b
.
The area under the curve is ln
a
√
13. Determine the area under y = 1 − x2 and above the x-axis. [It is much easier to use
geometry instead of calculus!]
Solution: The integral represents the area:
This is the quarter of the unit circle centered at (0,0). The unit circle has area π, so this
π
region has area .
4