The area of the region bounded by y-x=2 and x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given curves are $y = x+2$ and $y = x^2$.To find the points of intersection,set $x^2 = x+2$.$x^2 - x - 2 = 0$$(x-2)(x+1) = 0$So,$x = 2$ and $x = -

Vedclass · Accepted Answer

$\frac{9}{2}$

Vedclass · Answer

(B) Given curves are $y = x+2$ and $y = x^2$.To find the points of intersection,set $x^2 = x+2$.$x^2 - x - 2 = 0$$(x-2)(x+1) = 0$So,$x = 2$ and $x = -1$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 2$:$A = \int_{-1}^{2} (x+2 - x^2) \, dx$$A = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2}$$A = \left( \frac{4}{2} + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 - \frac{-1}{3} \right)$$A = \left( 2 + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right)$$A = \left( 6 - \frac{8}{3} \right) - \left( \frac{3 - 12 + 2}{6} \right)$$A = \frac{10}{3} - \left( -\frac{7}{6} \right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$.

The area of the region bounded by $y-x=2$ and $x^{2}=y$ is equal to:

Similar Questions

The area bounded by the curves $y^2 - x = 0$ and $y - x^2 = 0$ is

The area of the region bounded by the parabola $(y-2)^{2}=(x-1)$,the tangent to it at the point whose ordinate is $3$,and the $x$-axis is:

The area of the region bounded by $\frac{x^2}{9}+\frac{y^2}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ is

The area of the region enclosed between the parabolas $y^{2}=2x-1$ and $y^{2}=4x-3$ is

The area enclosed by the curves $y=\ln(x+e^{2})$,$x=\ln(2/y)$ (which is $y=2e^{-x}$) and $x=\ln 2$,above the line $y=1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module