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

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(B) Given $f(x) = \int_{x}^{x^2} (t - 1) \, dt$.Using the Leibniz integral rule,we differentiate $f(x)$ with respect to $x$:$f'(x) = (x^2 - 1) \cdot \

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$30$

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(B) Given $f(x) = \int_{x}^{x^2} (t - 1) \, dt$.Using the Leibniz integral rule,we differentiate $f(x)$ with respect to $x$:$f'(x) = (x^2 - 1) \cdot \frac{d}{dx}(x^2) - (x - 1) \cdot \frac{d}{dx}(x)$$f'(x) = (x^2 - 1)(2x) - (x - 1)(1)$$f'(x) = 2x^3 - 2x - x + 1 = 2x^3 - 3x + 1$.We can factorize $f'(x)$ by observing that $x=1$ is a root:$f'(x) = (x - 1)(2x^2 + 2x - 1)$.For $x \in [1, 3]$,$f'(x) \ge 0$,so $f(x)$ is an increasing function.The maximum value occurs at the right endpoint,$x = 3$.$f(3) = \int_{3}^{9} (t - 1) \, dt = \left[ \frac{t^2}{2} - t \right]_{3}^{9}$$f(3) = \left( \frac{81}{2} - 9 \right) - \left( \frac{9}{2} - 3 \right) = \frac{72}{2} - 6 = 36 - 6 = 30$.

If $f(x) = \int_{x}^{x^2} (t - 1) \, dt$ for $1 \le x \le 3$,then the global maximum value of $f(x)$ is

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