int_0^k(sqrt{k}-sqrt{t})^2 , dt = | vedclass

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Question

(C) Expand the integrand: $(\sqrt{k}-\sqrt{t})^2 = k + t - 2\sqrt{k}\sqrt{t}$.Now,integrate term by term with respect to $t$ from $0$ to $k$:$\int_0^k

Vedclass · Accepted Answer

$\frac{k^2}{6}$

Vedclass · Answer

(C) Expand the integrand: $(\sqrt{k}-\sqrt{t})^2 = k + t - 2\sqrt{k}\sqrt{t}$.Now,integrate term by term with respect to $t$ from $0$ to $k$:$\int_0^k (k + t - 2\sqrt{k}t^{1/2}) \, dt = \left[ kt + \frac{t^2}{2} - 2\sqrt{k} \cdot \frac{t^{3/2}}{3/2} \right]_0^k$$= \left[ kt + \frac{t^2}{2} - \frac{4}{3}\sqrt{k}t^{3/2} \right]_0^k$Substitute the limits:$= (k(k) + \frac{k^2}{2} - \frac{4}{3}\sqrt{k}(k^{3/2})) - (0)$$= k^2 + \frac{k^2}{2} - \frac{4}{3}k^2$$= \frac{3k^2}{2} - \frac{4k^2}{3} = \frac{9k^2 - 8k^2}{6} = \frac{k^2}{6}$.Thus,the correct option is $C$.

$\int_0^k(\sqrt{k}-\sqrt{t})^2 \, dt =$

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