JEE Main 2019 (Online) 9th April Morning ShiftQuestion 1
The area (in sq. units) of the region $A=\left\{(x,y):x^{2}\le y\le x+2\right\}$ is
Application of Integrals
Solution
$x^2=x+2\Rightarrow x=-1,2$. Area $=\int_{-1}^{2}(x+2-x^2)dx=\frac92$.Application of Integrals - JEE Mathematics question (JEE Main 2019 (Online) 9th April Morning Shift)
More from Application of Integrals
- The area bounded by the curves y=cos x and y=sin x between the ordinates x=0 and x=(3pi)/(2) is
- The area (in sq. units) of the region enclosed between the parabola y^2=2x and the line x+y=4 is…
- If alpha is the area of the common region that lies inside both the given ellipses, then the value of…
- The area (in square units) of the region enclosed by the ellipse x^2+3y^2=18 in the first quadrant below the…
- The area bounded by the curves y=|x^2-1| and y=1 is
- Sketch the curves and identify the region bounded by x=dfrac12 , x=2 , y=ln x and y=2^x . Find the area of…
- The area of the region bounded by the curve y=max\|x|,x|x-2|\ , the x -axis and the lines x=-2 and x=4 is…
- Area of the region \(x,y):x^2+(y-2)^2le4,\ x^2ge2y \ is