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Practice: Integration Technique (~F2017 Final Q27)
Related Topics
Wize University Calculus 1 Textbook > Integration Techniques
Integrating Even and Odd Functions
3 Activities
Practice: Integration Technique
Evaluate
∫
−
π
2
π
2
cos
2
x
sin
x
d
x
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2x\ \sin x\ dx
∫
−
2
π
2
π
cos
2
x
sin
x
d
x
.
0
0
0
1
1
1
−
1
-1
−
1
π
2
\frac{\pi}{2}
2
π
π
2
\pi^2
π
2
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Check Submission
More Integrating Even and Odd Functions Questions:
Practice: Integration Technique (~F2017 Final Q27)
Evaluate
∫
−
π
2
π
2
cos
2
x
sin
x
d
x
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2x\ \sin x\ dx
∫
−
2
π
2
π
cos
2
x
sin
x
d
x
.
Practice: Integration Technique (~F2017 Final Q27)
Practice: Integration Technique
Evaluate
∫
−
π
2
π
2
cos
2
x
sin
x
d
x
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2x\ \sin x\ dx
∫
−
2
π
2
π
cos
2
x
sin
x
d
x
.
Property of Definite Integral
Suppose
f
(
−
x
)
=
−
f
(
x
)
f(-x)=-f(x)
f
(
−
x
)
=
−
f
(
x
)
and
g
(
−
x
)
=
g
(
x
)
g\left(-x\right)=g\left(x\right)
g
(
−
x
)
=
g
(
x
)
for all
x
x
x
values.
If
∫
0
1
f
(
x
)
d
x
=
5
and
∫
−
1
0
g
(
x
)
d
x
=
7
\displaystyle\int_0^1f\left(x\right)dx=5\ \ \text{and}\ \ \int_{-1}^0g\left(x\right)dx=7
∫
0
1
f
(
x
)
d
x
=
5
and
∫
−
1
0
g
(
x
)
d
x
=
7
, evaluate
∫
−
1
1
[
2
f
(
x
)
+
3
g
(
x
)
]
d
x
\displaystyle \int_{-1}^1\left[2f\left(x\right)+3g\left(x\right)\right]dx
∫
−
1
1
[
2
f
(
x
)
+
3
g
(
x
)
]
d
x
Integrating Even and Odd Functions
Evaluate
∫
−
π
2
π
2
cos
2
x
sin
x
d
x
\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2x\sin x \ dx
∫
−
2
π
2
π
cos
2
x
sin
x
d
x
Evaluate
∫
−
7
π
3
7
π
3
sin
x
d
x
\displaystyle\int_{-\frac{7\pi}{3}}^{\frac{7\pi}{3}}\sin x\ dx
∫
−
3
7
π
3
7
π
sin
x
d
x
Evaluate
∫
−
70
70
x
3
+
x
5
d
x
\text{Evaluate} \displaystyle \int_{-70}^{70} x^3+x^5\space dx
Evaluate
∫
−
70
70
x
3
+
x
5
d
x
Integrating Even or Odd Functions
Evaluate
∫
−
π
π
(
2
sin
3
x
−
4
cos
x
)
d
x
.
\int_{-\pi}^{\pi} (2\sin^3x - 4\cos x)\ dx.
∫
−
π
π
(
2
sin
3
x
−
4
cos
x
)
d
x
.
Integrating Even and Odd Functions
Evaluate
∫
−
π
2
π
2
cos
2
x
sin
x
d
x
\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2x\sin x \ dx
∫
−
2
π
2
π
cos
2
x
sin
x
d
x
Suppose that
f
(
x
)
f(x)
f
(
x
)
is even,
∫
0
6
f
(
x
)
d
x
=
9
\int_{0}^{6}f(x)dx=9
∫
0
6
f
(
x
)
d
x
=
9
and
∫
−
3
3
f
(
x
)
d
x
=
6
\int_{-3}^{3}f(x)dx=6
∫
−
3
3
f
(
x
)
d
x
=
6
and
g
(
x
)
g(x)
g
(
x
)
is odd with
∫
−
3
6
g
(
x
)
d
x
=
3
\int_{-3}^{6}g(x)dx=3
∫
−
3
6
g
(
x
)
d
x
=
3
. What is the value of the integral
∫
3
6
2
f
(
x
)
−
3
g
(
x
)
d
x
\int_{3}^{6}2f(x)-3g(x)dx
∫
3
6
2
f
(
x
)
−
3
g
(
x
)
d
x
?
Evaluate
∫
−
7
π
3
7
π
3
sin
x
d
x
\displaystyle\int_{-\frac{7\pi}{3}}^{\frac{7\pi}{3}}\sin x\ dx
∫
−
3
7
π
3
7
π
sin
x
d
x
Integrating Even and Odd Functions: Definite Integrals
Evaluate
∫
−
1
1
4
−
(
x
+
1
)
2
+
sin
x
⋅
e
2
x
4
d
x
\int_{-1}^1\sqrt{4-\left(x+1\right)^2}+\sin x\cdot e^{2x^4}\ dx
∫
−
1
1
4
−
(
x
+
1
)
2
+
sin
x
⋅
e
2
x
4
d
x
.
Integrating Even and Odd Functions
Suppose that
f
(
−
x
)
=
f
(
x
)
f\left(-x\right)=f\left(x\right)
f
(
−
x
)
=
f
(
x
)
and
∫
0
5
5
⋅
f
(
x
)
d
x
=
9
\int_0^55\cdot f\left(x\right)dx=9
∫
0
5
5
⋅
f
(
x
)
d
x
=
9
. Find
∫
−
5
5
3
⋅
f
(
x
)
d
x
\int_{-5}^53\cdot f\left(x\right)\ dx
∫
−
5
5
3
⋅
f
(
x
)
d
x
.
Property of Definite Integral
Suppose
f
(
−
x
)
=
−
f
(
x
)
f(-x)=-f(x)
f
(
−
x
)
=
−
f
(
x
)
and
g
(
−
x
)
=
g
(
x
)
g\left(-x\right)=g\left(x\right)
g
(
−
x
)
=
g
(
x
)
for all
x
x
x
values.
If
∫
0
1
f
(
x
)
d
x
=
5
and
∫
−
1
0
g
(
x
)
d
x
=
7
\displaystyle\int_0^1f\left(x\right)dx=5\ \ \text{and}\ \ \int_{-1}^0g\left(x\right)dx=7
∫
0
1
f
(
x
)
d
x
=
5
and
∫
−
1
0
g
(
x
)
d
x
=
7
, evaluate
∫
−
1
1
[
2
f
(
x
)
+
3
g
(
x
)
]
d
x
\displaystyle \int_{-1}^1\left[2f\left(x\right)+3g\left(x\right)\right]dx
∫
−
1
1
[
2
f
(
x
)
+
3
g
(
x
)
]
d
x