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If $y=f(x)$ makes $+v e$ intercept of 2 and 0 unit on $x$ and $y$ axes and encloses an area of $3 / 4$ square unit with the axes then $\int_0^2 x f^{\prime}(x) d x ~~is$
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-3/4
-3/4
We have $\int_0^2 f(x) d x=\frac{3}{4} ;$ Now, $\int_0^2 x f^{\prime}(x) d x=x \int_0^2 f^{\prime}(x) d x-\int_0^2 f(x) d x$
$=[x f(x)]_0^2-\frac{3}{4}=2 f(2)-\frac{3}{4}=0-\frac{3}{4}(\because f(2)=0)=-\frac{3}{4}$
$=[x f(x)]_0^2-\frac{3}{4}=2 f(2)-\frac{3}{4}=0-\frac{3}{4}(\because f(2)=0)=-\frac{3}{4}$
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