Question

Evaluate the integral
${\int }_{-1}^1(\sqrt{1-x+x^2}-\sqrt{1+x+x^2})dx$

• A. $\frac{1}{2}$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

The correct option is C $0$

${\int }_{-1}^1(\sqrt{1-x+x^2}+\sqrt{1+x+x^2})dx=2{\int }_0^1(\sqrt{1-x+x^2}+\sqrt{1+x+x^2})dx$ (since function is symmetric about $x=0$ i.e. $f\left(x\right)=f(-x)$

$=2{\int }_0^1(\sqrt{(x^2-x+0.25)+0.75}+\sqrt{(x^2+x+0.25)+0.75})dx$

$=2{\int }_0^1(\sqrt{(x-0.5{)}^2+0.75}+\sqrt{(x+0.5{)}^2+0.75})dx$

$=2{(\frac{x-0.5}{2}\sqrt{(x-0.5{)}^2}+\frac{0.75}{2}\log |x-0.5+\sqrt{(x-0.5{)}^2+0.75}|+\frac{x+0.5}{2}\sqrt{(x+0.5{)}^2}+\frac{0.75}{2}\log }^{}{|x+0.5+\sqrt{(x+0.5{)}^2+0.75}|)}_0^1$

$=0$

This is required answer.