Area Under Curves Question 7
Q7 - 30 January - Shift 1
Let $\alpha$ be the area of the larger region bounded by the curve $y^{2}=8 x$ and the lines $y=x$ and $x=2$, which lies in the first quadrant. Then the value of $3 \alpha$ is equal to
Show Answer
Answer: 22
Solution:
Formula: Area between two curves - Area bounded by two intersecting curves and lines parallel to $\mathrm{y}-$ axis
$y=x$ and $ y^{2}=8 x$
Solving it
$ \begin{aligned} & x^{2}=8 x \\ \therefore \quad x & =0,8 \\ \text{ } & x=2 \text{ will intersect occur at } \\ y^{2} & =16 \Rightarrow \text{ } y= \pm 4 \text{ } \end{aligned} $
$\therefore$ Area of shaded $ =\int_2^{8}(\sqrt{8 x}-x) d x=\int_2^{8}(2 \sqrt{2} \sqrt{x}-x) d x $
Area of shaded $=[2 \sqrt{2} \cdot \frac{x^{3 / 2}}{3 / 2}-\frac{x^{2}}{2}]_2^{8} $
Area of shaded $=(\frac{4 \sqrt{2}}{3} \cdot 2^{9 / 2}-32)-(\frac{4 \sqrt{2}}{3} \cdot 2^{1 / 2}-2)$
Area of shaded $=[2^5 (\frac{4}{3}-1)-2(\frac{8}{3}-1)] = \frac{(32-10)}{3}$
Area of shaded $=\frac{22}{3}$
$A =\frac{22}{3}$
$3A =22$
