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Find the average value of the function f()=tan over the interval [0,(π)/4]
Related Topics
Wize University Calculus 1 Textbook > Applications of Integration
Average Value of a Function
3 Activities
Find the average value of the function
f
(
θ
)
=
sec
θ
tan
θ
f\left(\theta\right)=\sec\theta\tan\theta
f
(
θ
)
=
sec
θ
tan
θ
over the interval
[
0
,
π
4
]
\left[0,\dfrac{\pi}{4}\right]
[
0
,
4
π
]
Answer
I don't know
Check Submission
More Average Value of a Function Questions:
Average Value of Polynomial and Exponential Functions
Find the average value of the function
g
(
x
)
=
a
x
+
x
a
g\left(x\right)=a^x+x^a
g
(
x
)
=
a
x
+
x
a
over the interval
[
2
,
3
]
\left[2,3\right]
[
2
,
3
]
, where
a
≠
−
1
a\ \ne-1
a
=
−
1
is a constant.
Average Value of Polynomial and Exponential Functions
Find the average value of the function
g
(
x
)
=
a
x
+
x
a
g\left(x\right)=a^x+x^a
g
(
x
)
=
a
x
+
x
a
over the interval
[
2
,
3
]
\left[2,3\right]
[
2
,
3
]
, where
a
≠
−
1
a\ \ne-1
a
=
−
1
is a constant.
Average Value of a Function Problem
Find the average value of
y
=
x
2
y=x^2
y
=
x
2
over
[
1
,
3
]
[1, 3]
[
1
,
3
]
.
Find the average value of the function
f
(
x
)
=
1
x
2
f\left(x\right)=\frac{1}{x^2}
f
(
x
)
=
x
2
1
over the interval
[
1
,
3
]
\left[1,\ 3\right]
[
1
,
3
]
Average Value of a Function
Find the average value of
f
(
x
)
=
sin
x
f(x)=\sin x
f
(
x
)
=
sin
x
on
[
0
,
π
/
2
]
[0,\pi/2]
[
0
,
π
/2
]
.
Average Value of a Function Problem
Find the average value of
y
=
x
2
y=x^2
y
=
x
2
over
[
1
,
3
]
[1, 3]
[
1
,
3
]
.
Find the average value of the function
f
(
θ
)
=
sec
θ
tan
θ
f\left(\theta\right)=\sec\theta\tan\theta
f
(
θ
)
=
sec
θ
tan
θ
over the interval
[
0
,
π
4
]
\left[0,\dfrac{\pi}{4}\right]
[
0
,
4
π
]
Find the average value of the function
f
(
θ
)
=
sec
θ
tan
θ
f\left(\theta\right)=\sec\theta\tan\theta
f
(
θ
)
=
sec
θ
tan
θ
over the interval
[
0
,
π
4
]
\left[0,\dfrac{\pi}{4}\right]
[
0
,
4
π
]
Average Value of Polynomial and Exponential Functions
Find the average value of the function
g
(
x
)
=
a
x
+
x
a
g\left(x\right)=a^x+x^a
g
(
x
)
=
a
x
+
x
a
over the interval
[
2
,
3
]
\left[2,3\right]
[
2
,
3
]
, where
a
≠
−
1
a\ \ne-1
a
=
−
1
is a constant.