The area bounded by the curves {y^2} = 8x and | vedclass

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

Question

(B) The given curves are ${y^2} = 8x$ and $y = x$. To find the points of intersection,substitute $y = x$ into ${y^2} = 8x$: ${x^2} = 8x \Rightarrow {x

Vedclass · Accepted Answer

$\frac{32}{3} 	ext{ sq. units}$

Vedclass · Answer

(B) The given curves are ${y^2} = 8x$ and $y = x$. To find the points of intersection,substitute $y = x$ into ${y^2} = 8x$: ${x^2} = 8x \Rightarrow {x^2} - 8x = 0 \Rightarrow x(x - 8) = 0$. Thus,the points of intersection are $x = 0$ and $x = 8$. The required area is given by the integral: $	ext{Area} = \int_{0}^{8} (\sqrt{8x} - x) \, dx = \int_{0}^{8} (2\sqrt{2}x^{1/2} - x) \, dx$. Evaluating the integral: $= \left[ 2\sqrt{2} \cdot \frac{x^{3/2}}{3/2} - \frac{x^2}{2} ight]_{0}^{8} = \left[ \frac{4\sqrt{2}}{3} x^{3/2} - \frac{x^2}{2} ight]_{0}^{8}$. Substituting the limits: $= \left( \frac{4\sqrt{2}}{3} (8)^{3/2} - \frac{8^2}{2} ight) - 0 = \left( \frac{4\sqrt{2}}{3} \cdot 16\sqrt{2} - 32 ight) = \frac{128}{3} - 32 = \frac{128 - 96}{3} = \frac{32}{3} 	ext{ sq. units}$.

The area bounded by the curves ${y^2} = 8x$ and $y = x$ is

Similar Questions

The area (in sq. units) of the region consisting of points $(x,y)$ on the $X-Y$ plane which satisfy $|x| \le 1 + |y|$ and $|y| \le 1$ is:

The area included between the parabolas $y^{2} = 5x$ and $x^{2} = 5y$ is

The area (in sq. units) of the region $\{(x, y) \in R^{2} | 4 x^{2} \leq y \leq 8 x+12\}$ is:

Find the area lying between the curves $y^{2}=2x$ and $y=x$.

If the area between $y=m x^{2}$ and $x=m y^{2}$ $(m>0)$ is $1/4$ sq units,then the value of $m$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module