Question

find the average value of the function of the given interval f(x)==sin^2xcos^3x

Answer 1

\[\large\rm f(x)=\sin^2x \cos^3x\]What is the given interval? :)

Answer 2

The average function value, maybe we'll call that \(\rm f(c)\), can be determined using the Mean Value Theorem for Integrals.

Answer 3

We can achieve our average this way,\[\large\rm \frac{1}{b-a}\int\limits_a^b \sin^2x \cos^3x~dx=f(c)\]Still need interval though.

Answer 4

\[\int\limits_{-\pi}^{\pi}\]

Answer 5

i am stuck at \[-1/2\pi[1/3\sin^3x-1/5\sin^5x) |-\pi \to \pi\]

Answer 6

im getting a 0 as my answer
something is wrong but i don't know which part

Answer 7

\[1/2\pi \int\limits_{-\pi}^{\pi} \sin^2x(1-\sin^2x)cosxdx \]
\[-1/2\pi \int\limits_{-\pi}^{\pi}u^2-u^4du\]

Answer 8

\[f \left( x \right)=\sin ^2x \cos ^3x\]
\[f \left( -x \right)=\sin ^2(-x)\cos ^3(-x)=\sin ^2x \cos ^3x=f(x)\]
so f(x) is an even function.
f(x)\[=2\int\limits_{0}^{\pi}\sin ^2x \cos ^3x dx\]

\[\int\limits_{a}^{b}\sin ^2x \left( 1-\sin ^2x \right)\cos x dx=\int\limits_{a}^{b}\left( \sin ^2x \cos x-\sin ^4x \cos x \right) dx\]
\[=\left( \frac{ \sin ^3x }{ 3 }-\frac{ \sin ^5x }{ 5 } \right)\]

Answer 9

here a=0,b=\[\pi\]