\( \int_{0}^{2}\displaystyle x\sqrt{2-x}\,dx \) is equal to :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{16\sqrt{2}}{15} \)
The property \( \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx \) with a = 2 moves the root onto x alone:
\( \displaystyle I=\int_{0}^{2}x\sqrt{2-x}\,dx=\int_{0}^{2}(2-x)\sqrt{x}\,dx \)
\( \displaystyle I=\int_{0}^{2}\left(2x^{1/2}-x^{3/2}\right)dx \)
\( \displaystyle =\left[\frac{4}{3}x^{3/2}-\frac{2}{5}x^{5/2}\right]_{0}^{2} \)
Now \( 2^{3/2}=2\sqrt{2} \) and \( 2^{5/2}=4\sqrt{2} \):
\( \displaystyle I=\frac{4}{3}(2\sqrt{2})-\frac{2}{5}(4\sqrt{2}) \)
\( \displaystyle =\frac{8\sqrt{2}}{3}-\frac{8\sqrt{2}}{5} \)
\( \displaystyle =\frac{40\sqrt{2}-24\sqrt{2}}{15} \)
\( \displaystyle =\frac{16\sqrt{2}}{15} \)
NCERT (2026-27 edition):
Some Properties of Definite Integrals
