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\(\int_0^k(\sqrt{k}-\sqrt{t})^2 d t=\)
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2581 Upvotes
Verified Answer
The correct answer is:
\(\frac{k^2}{6}\)
\(\begin{aligned}
\int_0^k(\sqrt{k}-\sqrt{t}) d t & =\int_0^k(k+t-2 \sqrt{k} \sqrt{t}) d t \\
& =\left(k t+\frac{t^2}{2}\right)_0^k-2 \sqrt{k}\left(\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)_0^k \\
& =\left(k \cdot k+\frac{k^2}{2}\right)-\frac{4 \sqrt{k}}{3}\left(k^{\frac{3}{2}}\right)_0^k \\
& =\frac{3 k^2}{2}-\frac{4 \sqrt{k}}{3} \cdot k^{3 / 2}=\frac{k^2}{6}
\end{aligned}\)
Hence, option (c) correct.
\int_0^k(\sqrt{k}-\sqrt{t}) 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.
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