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Q.
The area of the region bounded by the curves
$y=9 \,x^{2}$ and $y=5 x^{2}+4$ (in sq units) is
TS EAMCET 2015
Solution:
Intersecting point of $y=9 \,x^{2}$ and $y=5 \,x^{2}+4$ is given by
$9 x^{2}=5\, x^{2}+4$
$\Rightarrow 4\, x^{2}=4 $
$ \Rightarrow x^{2}=1 $
$\Rightarrow x=\pm 1$
$ \therefore y=9(\pm 1)^{2}=9$
Hence, intersecting points are $(\pm 1,9)$.
$\therefore $ Required area $=2 \int_{0}^{1}\left[\left(5 \,x^{2}+4\right)-\left(9 \,x^{2}\right)\right] d x $
$=2 \int_{0}^{1}\left(4-4 \,x^{2}\right) d x $
$=2\left[4 x-\frac{4\, x^{3}}{3}\right]_{0}^{1} $
$=2\left(4-\frac{4}{3}\right)-(0-0)=2 \times \frac{8}{3} $
$=\frac{16}{3} $ sq units
