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OpenStudy (anonymous):

integral from -10 to 10 (100-x^2)^3/2 dx

OpenStudy (turingtest):

I would first like to point out that this function is even, and we have a theorem that can help us here

OpenStudy (anonymous):

\[\int\limits_{-10}^{10}(100-x^{2})^{3/2}dx\]

OpenStudy (turingtest):

if \(f(x)\) is even we have that\[\int_{-a}^af(x)dx=2\int_0^af(x)dx\]this makes sense when you think about the symmetry of an even graph like x^2

OpenStudy (turingtest):

so\[2\int_0^{10}(100-x^2)^{3/2}dx\]have you tried a trig sub?

OpenStudy (anonymous):

okey continue please..idont understan how we integrate this?

OpenStudy (turingtest):

have you ever done a trigonometric substitution before?

OpenStudy (anonymous):

i forgot it.

OpenStudy (turingtest):

try\[x=10\sin\theta\implies dx=10\cos\theta d\theta\]and see how far you get

OpenStudy (anonymous):

i know it but how substitute it? could you write?

OpenStudy (turingtest):

\[2\int_0^{10}(100-x^2)^{3/2}dx\]\[x=10\sin\theta\implies dx=10\cos\theta d\theta\]so we change the bounds\[x=0\implies\theta=0\]\[x=10\implies\theta=\frac\pi2\]subbing this in we get\[2\int_0^{10}[100-(10\sin\theta)^2]^{3/2}(10\cos\theta)d\theta\]simplify...

OpenStudy (turingtest):

\[2\int_0^{\frac\pi2}[100-(10\sin\theta)^2]^{3/2}(10\cos\theta)d\theta\]\[=20\int_0^{\frac\pi2}(100-100\sin^2\theta)^{3/2}\cos\theta d\theta\]\[=20000\int_0^{\frac\pi2}(1-\sin^2\theta)^{3/2}\cos\theta d\theta\]\[=20000\int_{0}^{\frac\pi2}(\cos^2\theta)^{3/2}\cos\theta d\theta\]\[=20000\int_0^{\frac\pi2}\cos^4\theta d\theta\]

OpenStudy (turingtest):

then use the formula\[\cos^2\theta=\frac12[1+\cos(2\theta)]\]twice and integrate

OpenStudy (anonymous):

thank u so much

OpenStudy (turingtest):

welcome!

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