Mathematics

# If  $f\left( {3 - x} \right) = f\left( x \right)$  then  $\int_1^2 {xf\left( x \right)} {\rm{ dx}}$ equals-

$\frac{3}{2}\int_1^2 {f\left( x \right)} {\rm{ dx}}$

##### SOLUTION
$f \left(3-x\right)=f\left(x\right)$
Then, $\int _{ 1 }^{ 2 }{ x } f\left( x \right) dx$
$=\int _{ 1 }^{ 2 }{ \left( 2+1-x \right) } f\left( 3-x \right) dx$
$=\int _{ 1 }^{ 2 }{ \left( 3-x \right) } f\left( x \right) dx$
$\Rightarrow \int _{ 1 }^{ 2 }{ xf\left( x \right) } dx=\int _{ 1 }^{ 2 }{ 3f\left( x \right) } -\int _{ 1 }^{ 2 }{ xf\left( x \right) } dx$
$2\int _{ 1 }^{ 2 }{ xf\left( x \right) } dx=\int _{ 1 }^{ 2 }{ 3f\left( x \right) }$
$\int _{ 1 }^{ 2 }{ xf\left( x \right) } dx=\dfrac { 3 }{ 2 } \int _{ 1 }^{ 2 }{ f\left( x \right) } dx$
Hence, the answer is $\dfrac { 3 }{ 2 } \int _{ 1 }^{ 2 }{ f\left( x \right) } dx.$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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