int [f(x)] dx from 0 to 2pi, if f = pi^2 for 0<=x<=pi,

[jbstahley]

integral f(x)dx from 0 to 2pi, where f(x) = pi^2 if 0<=x<pi and 2pi^2-x^2 if pi^2<=x<=2pi

Help is very much appreciated.

~ Jessica

 

[arthur ohlsten]

If I understand this correctly

integral f[x]dx from 0 to 2pi
f[x]=pi^2 a constant

pi^2 Int from 0 to 2pi dx
pi^2[x] evaluated at [2pi.0]
[pi^2][2pi-0]
2 pi^3 answer

Arthur

 

[soroban]

Re: More Definite Integrals...

Hello, Jessica!

I assume that there is a typo in the problem . . .

\(\displaystyle \L\int^{\;\;\;\;2\pi}_0\)\(\displaystyle dx\;\) where: \(\displaystyle \:f(x) \:=\:\begin{Bmatrix}\pi^2 & \text{ if }0\,\leq x < \pi \\ 2\pi^2\,-\,x^2 & \text{ if }\pi \leq x \leq 2\pi \end{Bmatrix}\)

Just make two integrals . . .

\(\displaystyle \L I_1\;=\;\int^{\;\;\;\pi}_0\pi^2\,dx \;=\;\pi^2x\bigg]^{\pi}_0 \;=\;\pi^3\)

\(\displaystyle \L I_2\;=\;\int^{\;\;\;2\pi}_{\pi}\left(\pi^2\,-\,x^2\right)\,dx \;=\;\pi^2x\,-\,\frac{1}{3}x^3\bigg]^{2\pi}_{\pi}\)

. . \(\displaystyle \L= \;\left(4\pi^3\,-\,\frac{8}{3}\pi^3\right)\,-\,\left(2\pi^3\,-\,\frac{1}{3}\pi^3\right) \;=\;\frac{4}{3}\pi^3\,-\,\frac{5}{3}\pi^2\;=\;-\frac{1}{3}\pi^3\)


Answer: \(\displaystyle \L\:I_1\,+\,I_2\;=\;\pi^3\,-\,\frac{1}{3}\pi^3\;=\;\fbox{\frac{2}{3}\pi^3}\)