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If $f(x)=\int_{-1}^{x}|t|$ dt, then for any $x \geq 0, f(x)$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{2}\left(1+x^{2}\right)$
We have,
$\begin{aligned} f(x) &=\int_{-1}^{x} t\left|d t=\int_{-1}^{0}\right| t\left|d t+\int_{0}^{x}\right| t \mid d t \\ &=-\int_{-1}^{0} t d t+\int_{0}^{x} t d t \\ &=\left[-\frac{t^{2}}{2}\right]_{-1}^{0}+\left[\frac{t^{2}}{2}\right]_{0}^{x} \\ &=-\frac{1}{2}\left(0^{2}-(-1)^{2}\right)+\frac{1}{2}\left(x^{2}-0^{2}\right) \\ &=-\frac{1}{2}(-1)+\frac{1}{2}\left(x^{2}\right)=\frac{1}{2}\left(1+x^{2}\right) \end{aligned}$
$\begin{aligned} f(x) &=\int_{-1}^{x} t\left|d t=\int_{-1}^{0}\right| t\left|d t+\int_{0}^{x}\right| t \mid d t \\ &=-\int_{-1}^{0} t d t+\int_{0}^{x} t d t \\ &=\left[-\frac{t^{2}}{2}\right]_{-1}^{0}+\left[\frac{t^{2}}{2}\right]_{0}^{x} \\ &=-\frac{1}{2}\left(0^{2}-(-1)^{2}\right)+\frac{1}{2}\left(x^{2}-0^{2}\right) \\ &=-\frac{1}{2}(-1)+\frac{1}{2}\left(x^{2}\right)=\frac{1}{2}\left(1+x^{2}\right) \end{aligned}$
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