Q96 / 155 · WBJEE 2017 · Mathematics
If $f(x)=\int_{-1}^{x}|t|$ dt, then for any $x \geq 0, f(x)$ is equal to
- A$\frac{1}{2}\left(1-x^{2}\right)$
- B$1-x^{2}$
- Correct answerC$\frac{1}{2}\left(1+x^{2}\right)$
- D$1+x^{2}$
Solution
We have,
$\begin{aligned} f(x) &=\int_{-1}^{x} t\left|d t=\int_{-1}^{0}\right| t\left|d t+\int_{0}^{x}\right| t \mid d t \\ &=-\int_{-1}^{0} t d t+\int_{0}^{x} t d t \\ &=\left[-\frac{t^{2}}{2}\right]_{-1}^{0}+\left[\frac{t^{2}}{2}\right]_{0}^{x} \\ &=-\frac{1}{2}\left(0^{2}-(-1)^{2}\right)+\frac{1}{2}\left(x^{2}-0^{2}\right) \\ &=-\frac{1}{2}(-1)+\frac{1}{2}\left(x^{2}\right)=\frac{1}{2}\left(1+x^{2}\right) \end{aligned}$
Hence, the correct answer is option: 'C'.