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

Question

Aicharya · Accepted Answer

$\begin{aligned} \int_0^k(\sqrt{k}-\sqrt{t}) d t & =\int_0^k(k+t-2 \sqrt{k} \sqrt{t}) d t \ & =\left(k t+\frac{t^2}{2}\right)_0^k-2 \sqrt{k}\left(\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)_0^k \ & =\left(k \cdot k+\frac{k^2}{2}\right)-\frac{4 \sqrt{k}}{3}\left(k^{\frac{3}{2}}\right)_0^k \ & =\frac{3 k^2}{2}-\frac{4 \sqrt{k}}{3} \cdot k^{3 / 2}=\frac{k^2}{6} \end{aligned}$ Hence, option (c) correct.

\(\int_0^k(\sqrt{k}-\sqrt{t})^2 d t=\)

Solution