Area Ques 65
Question
- Find the area bounded by the curve $x^{2}=4 y$ and the straight line $x=4 y-2$.
Show Answer
Answer:
Correct Answer: 65.$(\frac{9}{8} \text { sq units })$
Solution:
Formula:
- The point of intersection of the curves $x^{2}=4 y$ and $x=4 y-2$ could be sketched are $x=-1$ and $x=2$.
$\therefore$ Required area
$ \begin{aligned} & =\int _{-1}^{2} [(\frac{x+2}{4})-(\frac{x^{2}}{4})] d x \\ & =\frac{1}{4} [\frac{x^{2}}{2}+2 x-\frac{x^{3}}{3}] _{-1}^2 \\ & =\frac{1}{4} [(2+4-\frac{8}{3})-(\frac{1}{2}-2+\frac{1}{3})] \\ & =\frac{1}{4} [\frac{10}{3}-(\frac{-7}{6})]=\frac{1}{4} \cdot \frac{9}{2}=\frac{9}{8} \text { sq units } \end{aligned} $
BETA
