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

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

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

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

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

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