The value of int_0^1 4 x^3 left{ frac{d^2}{d | vedclass

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(B) Let $I = \int_0^1 4 x^3 \frac{d^2}{d x^2} (1-x^2)^5 d x$.Using integration by parts,let $u = 4x^3$ and $dv = \frac{d^2}{dx^2}(1-x^2)^5 dx$.Then $d

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$2$

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(B) Let $I = \int_0^1 4 x^3 \frac{d^2}{d x^2} (1-x^2)^5 d x$.Using integration by parts,let $u = 4x^3$ and $dv = \frac{d^2}{dx^2}(1-x^2)^5 dx$.Then $du = 12x^2 dx$ and $v = \frac{d}{dx}(1-x^2)^5 = 5(1-x^2)^4(-2x) = -10x(1-x^2)^4$.$I = [4x^3 \cdot (-10x(1-x^2)^4)]_0^1 - \int_0^1 (-10x(1-x^2)^4) \cdot 12x^2 dx$.The boundary term $[ -40x^4(1-x^2)^4 ]_0^1 = 0 - 0 = 0$.So,$I = 120 \int_0^1 x^3(1-x^2)^4 dx$.Let $t = 1-x^2$,then $dt = -2x dx$,so $x^2 = 1-t$ and $x dx = -\frac{1}{2} dt$.$I = 120 \int_1^0 (1-t) t^4 (-\frac{1}{2} dt) = 60 \int_0^1 (t^4 - t^5) dt$.$I = 60 [\frac{t^5}{5} - \frac{t^6}{6}]_0^1 = 60 (\frac{1}{5} - \frac{1}{6}) = 60 (\frac{6-5}{30}) = 60 (\frac{1}{30}) = 2$.

The value of $\int_0^1 4 x^3 \left\{ \frac{d^2}{d x^2} (1-x^2)^5 \right\} d x$ is

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