Practicing Success
Practicing Success
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If f(x) is differentibale and $\int_0^{t^2} xf(x)dx =\frac{2}{5}t^5$ then $f(\frac{4}{25})$ equals |
2/5 -5/2 1 5/2 |
2/5 |
$\int_0^{t^2} xf(x)dx =\frac{2}{5}t^5$ differentiate both side w.r.t. t $t^2\, f(t^2)\, 2t =\frac{10t^4}{5}⇒f(t^2) = t$ put $t = 2/5, f(4/ 25) = 2/5$ |
