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  4. Assertion: The area bounded by the curves y2=4 a2(x-1) and lines x=1 and y=4 a is (16 a/3) sq. units. Reason: The area enclosed between the parabola y2=x2-x+2 and the line y=x+2 is (8/3) sq. units.

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Q. Assertion : The area bounded by the curves $y^{2}=4 a^{2}(x-1)$ and lines $x=1$ and $y=4 a$ is $\frac{16 a}{3}$ sq. units.
Reason : The area enclosed between the parabola $y^{2}=x^{2}-x+2$ and the line $y=x+2$ is $\frac{8}{3}$ sq. units.

Application of Integrals

Solution:

Assertion : On solving $y ^{2}=4 a ^{2}( x -1)$ and $y =4 a$, we get $x=5$
image
Required area $=\int\limits_{1}^{5}(4 a -2 a \sqrt{ x -1}) dx $
$=\left[4 ax -2 a \cdot \frac{( x -1)^{3 / 2}}{3 / 2}\right]_{1}^{5}=\frac{16 a }{3} sq$.units
Reason : Given equation of parabola can be rewritten as
image
$\left(x-\frac{1}{2}\right)^{2}=y-\frac{7}{4}$
Parabola $y^{2}=x^{2}-x+2$ and the line $y=x+2$ intersects at the point $(0,0)$ and $(2,4)$
$\therefore $ Required area $=\int\limits_{0}^{2}\left[( x +2)-\left( x ^{2}- x +2\right)\right] dx$
$=\int\limits_{0}^{2}\left(-x^{2}+2 x\right) d x=\left[\frac{-x^{3}}{3}+x^{2}\right]_{0}^{2}=\frac{4}{3}$ sq.units