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(A) Given $F(x) = \int_0^x f(t) dt$,by the Fundamental Theorem of Calculus,$F'(x) = f(x)$ and $F(0) = 0$.We are given the integral equation:$\int_0^\p

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$4$

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(A) Given $F(x) = \int_0^x f(t) dt$,by the Fundamental Theorem of Calculus,$F'(x) = f(x)$ and $F(0) = 0$.We are given the integral equation:$\int_0^\pi (f'(x) + F(x)) \cos x dx = 2$Split the integral into two parts:$\int_0^\pi f'(x) \cos x dx + \int_0^\pi F(x) \cos x dx = 2$Apply integration by parts to the first integral $I_1 = \int_0^\pi f'(x) \cos x dx$:$I_1 = [f(x) \cos x]_0^\pi - \int_0^\pi f(x) (-\sin x) dx$$I_1 = f(\pi) \cos(\pi) - f(0) \cos(0) + \int_0^\pi f(x) \sin x dx$Since $f(\pi) = -6$,$\cos(\pi) = -1$,and $\cos(0) = 1$:$I_1 = (-6)(-1) - f(0)(1) + \int_0^\pi f(x) \sin x dx = 6 - f(0) + \int_0^\pi f(x) \sin x dx$Now,consider the second integral $I_2 = \int_0^\pi F(x) \cos x dx$. Apply integration by parts:$I_2 = [F(x) \sin x]_0^\pi - \int_0^\pi F'(x) \sin x dx$Since $F(0) = 0$ and $\sin(0) = \sin(\pi) = 0$,the boundary term is $0$.$I_2 = - \int_0^\pi f(x) \sin x dx$Substitute $I_1$ and $I_2$ back into the original equation:$(6 - f(0) + \int_0^\pi f(x) \sin x dx) + (- \int_0^\pi f(x) \sin x dx) = 2$$6 - f(0) = 2$$f(0) = 4$.

Let $f: R \rightarrow R$ be a differentiable function such that its derivative $f^{\prime}$ is continuous and $f(\pi)=-6$. If $F:[0, \pi] \rightarrow R$ is defined by $F(x)=\int_0^{ x } f( t ) dt$,and if $\int_0^\pi\left(f^{\prime}( x )+ F ( x )\right) \cos x dx =2$,then the value of $f(0)$ is.

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