Area Question 53
Question 53
- Find the area bounded by the curve $x^{2}=4 y$ and the straight line $x=4 y-2$.
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Solution:
- 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
$$ \begin{aligned} & =\int_{-1}^{2} \frac{x+2}{4}-\frac{x^{2}}{4} d x \ & =\frac{1}{4} \frac{x^{2}}{2}+2 x-\frac{x^{3}}{3} \ & =\frac{1}{4} 2+4-\frac{8}{3}-\frac{1}{2}-2+\frac{1}{3} \ & =\frac{1}{4} \frac{10}{3}-\frac{-7}{6}=\frac{1}{4} \cdot \frac{9}{2}=\frac{9}{8} \text { sq units } \end{aligned} $$
