Question

Evaluate ${\int }_3^{7}x\sqrt{x^2-3}dx$

Answer 1

We have,

$I={\int }_3^{7}x\sqrt{x^2-3}dx$

Let $t=x^2-3$

$\frac{dt}{dx}=2x-0$

$\frac{dt}{2}=xdx$

Therefore,

$I=\frac{1}{2}{\int }_6^{46}\sqrt{t}dt$

$I=\frac{1}{2}{\left[\frac{t^{3/2}}{\frac{3}{2}}\right]}_6^{46}$

$I=\frac{1}{3}[{\left(46\right)}^{3/2}-{\left(6\right)}^{3/2}]$

Hence, this is the answer.