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If the line, $y=mx$ , bisects the area of the region

$\left \{ \left ( x,y \right ):0\leq x\leq \frac{3}{2},0\leq y\leq 1+4x-x^{2} \right \},$ then $m$ equals :

Option 1)
$\frac{39}{16}$

Option 2)
$\frac{9}{8}$

Option 3)
$\frac{13}{3}$

Option 4)
$\frac{13}{6}$

Answers (1)

best_answer

Geometrical interpretation of a definite integral -

An algebraic sum of the area of the figure bounded by the curve $y = f(x),$ the x axis and the striaght lines x=a and x=b. The areas above x axis are taken as positive and the areas below x axis are taken as negative.

- wherein

Where $a< b$

Hence

$\int_{a}^{b}f\left ( x \right )dx=$

$ar\left ( PAQP \right )-ar\left ( QTRQ \right )+ar\left ( RBSR \right )$

$A_1=\frac{1}{2}*\frac{3}{2}*\frac{3m}{2}= \frac{9m}{8}$

Total Area $=\int_{0}^{3/2}\left ( -x^2+4x+1 \right )$

$=\left [ \frac{-x^3}{3}+2x^2+x \right ]_{0}^{3/2}$

$=\frac{-9}{8}+\frac{9}{2}+\frac{3}{2}= \frac{39}{8}$

So, $\frac{39}{8}=2*\frac{9m}{8}$

$\Rightarrow m= \frac{13}{6}$

Option 1)

$\frac{39}{16}$

Option 2)

$\frac{9}{8}$

Option 3)

$\frac{13}{3}$

Option 4)

$\frac{13}{6}$

Posted by

prateek

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