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 (in sq. units) bounded by the curve $y=2x-x^2$ and the line $y=-x$ is

Let the area of the region $\{(x, y): 0 \leq x \leq 3, 0 \leq y \leq \min \{x^2+2, 2x+2\}\}$ be $A$. Then $12A$ is equal to

If the area bounded by the curves $y=ax^2$ and $x=ay^2$ $(a>0)$ is $3$ sq units,then the value of $a$ is

Let the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=e^{x-1}-e^{-|x-1|}$ and $g(x)=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right)$. Then the area of the region in the first quadrant bounded by the curves $y=f(x)$,$y=g(x)$ and $x=0$ is

The area enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ over the interval $\left[0, \frac{\pi}{2}\right]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module