Using properties of definite integrals, prove that $\int_{0}^{2} x \sqrt{2-x}\, dx=\frac{16 \sqrt{2}}{15}$

CBSE Class 12 · 2007 · Definite IntegralsVery Short Answer

Using properties of definite integrals, prove that

\int_{0}^{2} x 2 - x d x = \frac{16 2}{15}

Answer: Proved

Let

I = \int_{0}^{2} x 2 - x d x

. Apply

\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x

with

a = 2

(replace

x \to 2 - x

I = \int_{0}^{2} (2 - x) x d x = \int_{0}^{2} (2 x^{1/2} - x^{3/2}) d x

= [\frac{4}{3} x^{3/2} - \frac{2}{5} x^{5/2}]_{0}^{2}

With

2^{3/2} = 22

and

2^{5/2} = 42

= \frac{4}{3} (22) - \frac{2}{5} (42) = \frac{8 2}{3} - \frac{8 2}{5} = 82 \cdot \frac{5 - 3}{15} = \frac{16 2}{15}

Hence proved.

CBSE Class 12 Board · 2007 · Subjective