I think your lower end point could be possible x = 0 and your upper endpoint being x =2 so \[\Delta x= (b-a)/n\]
OpenStudy (across):
just set it up \[\int_{0}^{2}\int_{0}^{\sin^{-1}\left(x/2\right)}dydx\]
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ganeshie8 (ganeshie8):
riemann sum's are no fun @sweetburger
OpenStudy (sweetburger):
this might be above me. ill see myself out
OpenStudy (arianna1453):
Im confused
OpenStudy (across):
Do you know how to perform the integral that I showed you above? Hint:
\[\int_{0}^{2}\int_{0}^{\sin^{-1}\left(x/2\right)}dydx=\int_{0}^{2}\sin^{-1}\left(\frac x2\right)\,dx.\]
OpenStudy (sweetburger):
Didn't know that could be rewritten like that . Starting to make some sense to me at least.
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OpenStudy (arianna1453):
Not with this question
OpenStudy (across):
-_-
Then ask a new question and close this one.
ganeshie8 (ganeshie8):
question explicitly asks to integrate with respect to \(y\) and I don't think op knows double integrals...
OpenStudy (across):
Sure. Then turn it into a type 2 integral:\[\int_{0}^{\sin^{-1}\left(1\right)}2-2\sin\left(y\right)\,dy.\]