Question

If $f\left(x\right)=\underset{-1}{\overset{x}{\int }}|t|dt$, 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 C $\frac{1}{2}(1+x^2)$

$f\left(x\right)=\underset{-1}{\overset{x}{\int }}|t|dt=\underset{-1}{\overset{0}{\int }}-tdt+\underset{0}{\overset{x}{\int }}tdt\Rightarrow f\left(x\right)=\frac{1}{2}+\frac{x^2}{2}$