Find the value of \(\rm \int_0^1 x \sqrt{x^2 + 4}\;dx\)

Question

Question

Answer 1

Concept:

\(\rm \int x^n dx = \frac{x^{n+1}}{n+1}+c\)

Calculation:

I = \(\rm \int_0^1 x \sqrt{x^2 + 4}\;dx\)

Let x² + 4 = t

Differentiating with respect to x, we get

⇒ 2xdx = dt

⇒ xdx = \(\rm \frac {dt}{2}\)

x	0	1
t	4	5

Now,

I = \(\rm \frac{1}{2}\int_4^5 \sqrt{t}\;dt\)

= \(\rm \frac{1}{2} \left[\frac{t^{3/2}}{\frac{3}{2}} \right ]_4^5\)

= \(\rm \frac{1}{3} \left[5^{3/2} - 4^{3/2} \right ]\)

= \(\rm \frac{1}{3}[5\sqrt 5 - 8]\)