Let f: R R be a differentiable function such that its derivative … - Vidyadip

Question

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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Answer

a
$F ( x )=\int_0^{ x } f( t ) \cdot dt$

$\Rightarrow F ^{\prime}( x )=f( x )$

$I =\int_0^\pi f^{\prime}( x ) \cdot \cos x d x +\int_0^\pi F ( x ) \cos ( x ) dx =2 .$ $. . . . . . (1)$

$I _1=\int_0^\pi f^{\prime}( x ) \cdot \cos xdx \text { (Let) }$

Using by parts

$I _1=(\cos x \cdot f( x ))_0^\pi+\int_0^\pi \sin x . f( x ) dx$

$I _1=6-f(0)+\int_0^\pi \sin x \cdot F ^{\prime}( x ) dx$

$I _1=6-f(0)+ I _2$ $. . . . . . (2)$

$I _2=\int_0^\pi \sin x \cdot F ^{\prime}( x ) \cdot dx$

Using by part we get

$I_2=(\sin x \cdot F ( x ))_0^\pi-\int_0^\pi \cos x F( x ) dx$

$I _2=-\int_0^\pi \cos x \cdot F ( x ) dx$

$(2)$ $\Rightarrow I _1=6-f(0)-\int_0^\pi \cos x \cdot F ( x ) dx$

$(1) \Rightarrow I =6-f(0)=2 \Rightarrow f(0)=4$

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