The area bounded by the curve y^2=2x+1 and th | vedclass

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

Question

(D) Given curves are $y^2 = 2x + 1$ and $x - y = 1$.From the line equation,$x = y + 1$.Substituting $x$ in the curve equation: $y^2 = 2(y + 1) + 1 \im

Vedclass · Accepted Answer

$\frac{16}{3}$ sq. units

Vedclass · Answer

(D) Given curves are $y^2 = 2x + 1$ and $x - y = 1$.From the line equation,$x = y + 1$.Substituting $x$ in the curve equation: $y^2 = 2(y + 1) + 1 \implies y^2 = 2y + 3 \implies y^2 - 2y - 3 = 0$.Factoring the quadratic: $(y - 3)(y + 1) = 0$,so $y = 3$ and $y = -1$.The area bounded is given by the integral with respect to $y$ from $-1$ to $3$ of the difference between the line and the curve: $x_{line} - x_{curve} = (y + 1) - \frac{y^2 - 1}{2}$.$	ext{Area} = \int_{-1}^3 \left( y + 1 - \frac{y^2 - 1}{2} ight) dy = \int_{-1}^3 \left( \frac{2y + 2 - y^2 + 1}{2} ight) dy = \frac{1}{2} \int_{-1}^3 (3 + 2y - y^2) dy$.$= \frac{1}{2} \left[ 3y + y^2 - \frac{y^3}{3} ight]_{-1}^3$.$= \frac{1}{2} \left[ (9 + 9 - 9) - (-3 + 1 + \frac{1}{3}) ight] = \frac{1}{2} \left[ 9 - (-\frac{5}{3}) ight] = \frac{1}{2} \left( \frac{27 + 5}{3} ight) = \frac{1}{2} 	imes \frac{32}{3} = \frac{16}{3} 	ext{ sq. units}$.

The area bounded by the curve $y^2=2x+1$ and the line $x-y=1$ is

Similar Questions

The area bounded by the curves $y=\frac{1}{4}\left|4-x^2\right|$ and $y=7-|x|$ is

Area of the region enclosed between the curves $x = y^2 - 1$ and $x = |y| \sqrt{1 - y^2}$ is

The area (in $sq. units$) of the region $A = \{(x,y) : \frac{y^2}{2} \le x \le y + 4\}$ is

The area (in sq. units) of the region lying in the first quadrant and enclosed by the $X$-axis,the straight line $x - \sqrt{3}y = 0$ and the circle $x^2 + y^2 = 4$ is

If the area lying in the first quadrant and bounded by the circle $x^2+y^2-4x=0$,the parabola $y^2=x$ and the $X$-axis is $A$,then $6A-9\sqrt{3}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module