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Q.
The line $y = mx$ bisects the area enclosed by the curve $y =1+4 x - x ^2 \&$ the lines $x =0, x =\frac{3}{2}$ $\& y=0$. Then the value of $m$ is :
Application of Integrals
Solution:
$2 \cdot \frac{1}{2}\left(\frac{3}{2} \cdot \frac{3 m }{2}\right)=\int\limits_0^{3 / 2}\left(1+4 x - x ^2\right) dx$
$\left.\frac{9 m }{4}= x +2 x ^2-\frac{ x ^3}{3}\right]_0^{3 / 2}=\frac{3}{2}+\frac{9}{2}-\frac{27}{8} \cdot \frac{1}{3}=6-\frac{9}{8}=\frac{39}{8}$
$ m =\frac{13}{6} $