If f(x) = int_{-1}^{x} |t| , dt, then for any | vedclass

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(C) Given $f(x) = \int_{-1}^{x} |t| \, dt$. For $x \geq 0$, we can split the integral at $t=0$: $f(x) = \int_{-1}^{0} |t| \, dt + \int_{0}^{x} |t| \,

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

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(C) Given $f(x) = \int_{-1}^{x} |t| \, dt$. For $x \geq 0$, we can split the integral at $t=0$: $f(x) = \int_{-1}^{0} |t| \, dt + \int_{0}^{x} |t| \, dt$. Since $|t| = -t$ for $t $f(x) = \int_{-1}^{0} (-t) \, dt + \int_{0}^{x} t \, dt$. Evaluating the integrals: $f(x) = \left[ -\frac{t^2}{2} \right]_{-1}^{0} + \left[ \frac{t^2}{2} \right]_{0}^{x}$. $f(x) = \left( 0 - (-\frac{(-1)^2}{2}) \right) + \left( \frac{x^2}{2} - 0 \right)$. $f(x) = \frac{1}{2} + \frac{x^2}{2} = \frac{1}{2}(1+x^2)$.

If $f(x) = \int_{-1}^{x} |t| \, dt$, then for any $x \geq 0$, $f(x)$ is equal to

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