If intlimits_0^{f(x)} {{t^2},dt} = x cos(pi x | vedclass

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(A) Given $\int\limits_0^{f(x)} {{t^2}\,dt} = x \cos(\pi x)$.Applying the Fundamental Theorem of Calculus,we evaluate the integral:$\left[ \frac{t^3}{

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is equal to $-\frac{1}{9}$

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(A) Given $\int\limits_0^{f(x)} {{t^2}\,dt} = x \cos(\pi x)$.Applying the Fundamental Theorem of Calculus,we evaluate the integral:$\left[ \frac{t^3}{3} \right]_0^{f(x)} = x \cos(\pi x)$$\Rightarrow \frac{[f(x)]^3}{3} = x \cos(\pi x)$$\Rightarrow [f(x)]^3 = 3x \cos(\pi x) \quad \dots(1)$At $x = 9$,$[f(9)]^3 = 3(9) \cos(9\pi) = 27(-1) = -27$.Thus,$f(9) = -3$.Differentiating both sides of equation $(1)$ with respect to $x$ using the chain rule and the product rule:$3[f(x)]^2 \cdot f'(x) = \frac{d}{dx} [3x \cos(\pi x)]$$3[f(x)]^2 \cdot f'(x) = 3 \cos(\pi x) - 3x \pi \sin(\pi x)$$[f(x)]^2 \cdot f'(x) = \cos(\pi x) - x \pi \sin(\pi x)$Substitute $x = 9$:$[f(9)]^2 \cdot f'(9) = \cos(9\pi) - 9\pi \sin(9\pi)$$(-3)^2 \cdot f'(9) = -1 - 9\pi(0)$$9 \cdot f'(9) = -1$$f'(9) = -\frac{1}{9}$.

If $\int\limits_0^{f(x)} {{t^2}\,dt} = x \cos(\pi x)$,then find $f'(9)$.

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