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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area of the region lying in the first quadrant and bounded by \(y = 4x^2, x = 0, y = 1\) and \(y = 4.\)

$\begin{array}{1 1} \frac{7}{3} \;sq.units \\ \frac{5}{3} \;sq.units \\ \frac{11}{3}\;sq.units \\ \frac{10}{3}\;sq.units \end{array} $

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Clearly the shaded region is the region lying in the first quadrant and bounded by $y=4x^2$,x=0,y=1 and y=4.
so the required area is given by
$A=\int_1^4xdy$,But $x=\sqrt{\frac{y}{4}}$
$\;\;\;=\sqrt{\frac{y}{4}}dy=\frac{1}{2}\int_1^4\sqrt ydy$
$\;\;\;=\frac{1}{2}\begin{bmatrix}\frac{y^{\frac{1}{2}+1}}{\frac{1}{2}+1}\end{bmatrix}^4=\begin{bmatrix}y^{\frac{3}{2}}\end{bmatrix}_1^4$
Applying limits:
$\;\;\;=\frac{1}{3}\begin{bmatrix}(4)^\frac{3}{2}-1^\frac{3}{2}\end{bmatrix}$
$\;\;\;=\frac{1}{3}\begin{bmatrix}(2^2)^\frac{3}{2}-(1)^\frac{3}{2}\end{bmatrix}=\frac{1}{3}\begin{bmatrix}(2)^3-1\end{bmatrix}$
$\;\;\;=\frac{1}{3}[8-1]$
$\;\;\;=\frac{1}{3}\times 7=\frac{7}{3}sq.units.$
Hence the required area is $\frac{7}{3}$sq.units.
answered Dec 21, 2013 by yamini.v
 

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