The value of the definite integral \(I = \int_{0}^{2}\displaystyle x\sqrt{2 - x}\,dx\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{16\sqrt{2}}{15}\)
By the property of definite integrals \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\) with a = 2, x is replaced by 2 − x:
\( \displaystyle I = \int_{0}^{2} (2 - x)\sqrt{x}\,dx \)
\( \displaystyle = \int_{0}^{2} \left(2x^{1/2} - x^{3/2}\right) dx \)
\( \displaystyle = \left[\frac{4}{3}x^{3/2} - \frac{2}{5}x^{5/2}\right]_{0}^{2} \)
With \(2^{3/2} = 2\sqrt{2}\) and \(2^{5/2} = 4\sqrt{2}\):
\( \displaystyle I = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5} \)
\( \displaystyle = 8\sqrt{2} \times \frac{5 - 3}{15} = \frac{16\sqrt{2}}{15} \)
NCERT (2026-27 edition):
Some Properties of Definite Integrals
