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across
 3 years ago
Best ResponseYou've already chosen the best response.1\[\int_{0}^{1}\pi(\sqrt[4]{x})^2dx\int_{0}^{1}\pi(x^4)^2dx\]

ta123
 3 years ago
Best ResponseYou've already chosen the best response.1how come but side are the same

ta123
 3 years ago
Best ResponseYou've already chosen the best response.1oh, wait my eyes are playing tricks on me I thought I saw a square root in the right integral

across
 3 years ago
Best ResponseYou've already chosen the best response.1Yeah, integrals will do that to your eyes. :P

ta123
 3 years ago
Best ResponseYou've already chosen the best response.1I got 7pi/9 is that what you got?

across
 3 years ago
Best ResponseYou've already chosen the best response.1\[\pi\int_{0}^{1}\sqrt{x}dx=\frac{2}{3}\pi\left[x^{\frac{3}{2}}\right]_{0}^{1}=\frac{2}{3}\pi,\]\[\pi\int_{0}^{1}x^8dx=\pi\left[\frac{1}{9}x^9\right]_{0}^{1}=\frac{1}{5}=\frac{1}{9}\pi,\]\[\frac{2}{3}\pi\frac{1}{9}\pi=\frac{5}{9}\pi.\]

ta123
 3 years ago
Best ResponseYou've already chosen the best response.1dang I integrated wrong by different number number to the exponents beside the number 1

ta123
 3 years ago
Best ResponseYou've already chosen the best response.1oops! I forgot "adding" in between by and different
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