Exam

JEE Main

Year

2019

Date

11 January

Shift

Morning Shift

Subject

Mathematics

Chapter

Area Under Curves

JEE Main 2019 · 11 January · Morning Shift · Mathematics · Area Under Curves

The area (in sq. units) of the region bounded by the curve x^ 2 =4...

The area (in sq. units) of the region bounded by the curve $x^{2}=4 y$ and the straight line $x=4 y-2$ is :

Step-by-step solution

Let points of intersection of the curve and the line be P and $Q x^{2}=4\left(\frac{x+2}{4}\right) x^{2}-x-2=0 x=2,-1$ Point are (2,1)$and \left(-1, \frac{1}{4}\right) \text{ Area }=\int_{-1}^{2}$\left[\left(\frac{x+2}{4}\right)-\left(\frac{x^{2}}{4}\right)\right] d x=\left[\frac{x^{2}}{8}+\frac{1}{2} x-\frac{x^{3}}{12}\right]_{-1}^{2} =\left(\frac{1}{2}+1-\frac{2}{3}\right)-\left(\frac{1}{8}-\frac{1}{2}+\frac{1}{12}\right)=\frac{9}{8}