If f(x) = int_{x}^{x^2} (t - 1) dt,1 le x le | vedclass

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(C) Given $f(x) = \int_{x}^{x^2} (t - 1) dt$. Using the Leibniz rule for differentiation,$f'(x) = (x^2 - 1) \cdot \frac{d}{dx}(x^2) - (x - 1) \cdot \f

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$4$

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(C) Given $f(x) = \int_{x}^{x^2} (t - 1) dt$. Using the Leibniz rule for differentiation,$f'(x) = (x^2 - 1) \cdot \frac{d}{dx}(x^2) - (x - 1) \cdot \frac{d}{dx}(x)$. $f'(x) = (x^2 - 1)(2x) - (x - 1)(1) = 2x^3 - 2x - x + 1 = 2x^3 - 3x + 1$. Factoring $f'(x)$,we get $f'(x) = (x - 1)(2x^2 + 2x - 1)$. For $1 \le x \le 2$,$f'(x) > 0$,which means $f(x)$ is a strictly increasing function on the interval $[1, 2]$. Since $f(x)$ is strictly increasing,the global maximum value occurs at the right endpoint,$x = 2$. $f(2) = \int_{2}^{4} (t - 1) dt = \left[ \frac{t^2}{2} - t \right]_{2}^{4} = (8 - 4) - (2 - 2) = 4 - 0 = 4$.

If $f(x) = \int_{x}^{x^2} (t - 1) dt$,$1 \le x \le 2$,then the global maximum value of $f(x)$ is

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