Question

If $f\left(x\right)={\int }_{-1}^x|t|dt$, then for any $x\ge 0,f\left(x\right)$ is equal to

• A. $\frac{1}{2}(1-x^2)$
• B. $1-x^2$
• C. $\frac{1}{2}(1+x^2)$
• D. $1+x^2$

Answer 1

The correct option is D $1+x^2$

$f\left(x\right)={\int }_{-1}^x|t|dt$

Since, $x\ge 0$

$\therefore {\int }_{-1}^x|t|dt={\int }_{-1}^0(-t)dt+{\int }_0^{x}tdt$

$=-{\left|\frac{t^2}{2}\right|}_{-1}^0+{\left|\frac{t^2}{2}\right|}_0^x$

$=\frac{1}{2}+\frac{x^2}{2}$

$=\frac{1}{2}(1+x^2)$.

Hence, for any $x\ge 0$, $f\left(x\right)=\frac{1}{2}(1+x^2)$