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(B) Given $F(x) = \int_0^{x^2} f(\sqrt{t}) dt$. Using the Leibniz rule for differentiation under the integral sign,we get $F'(x) = f(\sqrt{x^2}) \cdot

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$e^4-1$

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(B) Given $F(x) = \int_0^{x^2} f(\sqrt{t}) dt$. Using the Leibniz rule for differentiation under the integral sign,we get $F'(x) = f(\sqrt{x^2}) \cdot \frac{d}{dx}(x^2) = f(x) \cdot 2x$.Given $F'(x) = f'(x)$,we have $f'(x) = 2x f(x)$.This is a first-order linear differential equation: $\frac{f'(x)}{f(x)} = 2x$.Integrating both sides with respect to $x$,we get $\ln|f(x)| = x^2 + C$.Since $f(0) = 1$,we have $\ln(1) = 0^2 + C$,which implies $C = 0$.Thus,$\ln(f(x)) = x^2$,which gives $f(x) = e^{x^2}$.Now,$F(x) = \int_0^{x^2} e^{(\sqrt{t})^2} dt = \int_0^{x^2} e^t dt = [e^t]_0^{x^2} = e^{x^2} - e^0 = e^{x^2} - 1$.Therefore,$F(2) = e^{2^2} - 1 = e^4 - 1$.

Let $f:[0,2] \rightarrow R$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1$. Let $F(x)=\int_0^{x^2} f(\sqrt{t}) dt$ for $x \in [0,2]$. If $F'(x)=f'(x)$ for all $x \in (0,2)$,then $F(2)$ equals

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