The value of x that maximises the value of th | vedclass

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(C) Let $F(x) = \int\limits_x^{x + 3} {t(5 - t)\,dt}$.Using the Leibniz integral rule,we differentiate $F(x)$ with respect to $x$:$F'(x) = (x + 3)(5 -

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$1$

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(C) Let $F(x) = \int\limits_x^{x + 3} {t(5 - t)\,dt}$.Using the Leibniz integral rule,we differentiate $F(x)$ with respect to $x$:$F'(x) = (x + 3)(5 - (x + 3)) \cdot \frac{d}{dx}(x + 3) - x(5 - x) \cdot \frac{d}{dx}(x)$$F'(x) = (x + 3)(2 - x) - x(5 - x)$$F'(x) = (2x - x^2 + 6 - 3x) - (5x - x^2)$$F'(x) = 6 - x - x^2 - 5x + x^2$$F'(x) = 6 - 6x$.To find the critical points,set $F'(x) = 0$:$6 - 6x = 0 \Rightarrow x = 1$.Now,find the second derivative to check for maxima:$F''(x) = -6$.Since $F''(1) = -6 < 0$,the function $F(x)$ has a local maximum at $x = 1$.

The value of $x$ that maximises the value of the integral $\int\limits_x^{x + 3} {t(5 - t)\,dt}$ is

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