The value of \(\int_{0}^{2}\displaystyle x\sqrt{2-x}\,dx\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{16\sqrt{2}}{15}\)
By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx\) with \(a = 2\), replace x by \(2-x\):
\( \displaystyle \int_{0}^{2}x\sqrt{2-x}\,dx = \int_{0}^{2}(2-x)\sqrt{x}\,dx \)
\( \displaystyle = \int_{0}^{2}\left(2x^{1/2} - x^{3/2}\right)dx \)
\( \displaystyle = \left[\frac{4}{3}x^{3/2} - \frac{2}{5}x^{5/2}\right]_{0}^{2} \)
With \(2^{3/2} = 2\sqrt{2}\) and \(2^{5/2} = 4\sqrt{2}\):
\( \displaystyle = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5} \)
\( \displaystyle = \frac{40\sqrt{2} - 24\sqrt{2}}{15} = \frac{16\sqrt{2}}{15} \)