Mathematics 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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