Asked by Anonymous
can somebody show me how to get to the answer on this question? I try to solve it many time and counldn't get to the answer10π/21. the qeustion is πintegral from 0 to1 [((1-x^3)^2)- ((1-√x)^2)]dx
Answers
Answered by
Steve
Too bad you couldn't be bothered to show your work. It's just a bunch of straightforward power rule stuff.
∫[((1-x^3)^2)- ((1-√x)^2)] dx
= ∫[(1-2x^3+x^6)-(1-2√x+x)] dx
= ∫[-2x^3+x^6+2√x-x)] dx
= 1/7 x^7 - 1/2 x^4 - 1/2 x^2 + 4/3 x^(3/2)
so, evaluate that at 1 and 0, then multiply by π and you have
π(1/7 - 1/2 - 1/2 + 4/3) = 10π/21
∫[((1-x^3)^2)- ((1-√x)^2)] dx
= ∫[(1-2x^3+x^6)-(1-2√x+x)] dx
= ∫[-2x^3+x^6+2√x-x)] dx
= 1/7 x^7 - 1/2 x^4 - 1/2 x^2 + 4/3 x^(3/2)
so, evaluate that at 1 and 0, then multiply by π and you have
π(1/7 - 1/2 - 1/2 + 4/3) = 10π/21
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