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 of the region lying between the curves $y=\sqrt{4-x^2}$,$y^2=3x$ and the $Y$-axis is

Find the area of the region bounded by the parabola $(y-2)^2 = x-1$,the tangent to the parabola at the point $(2,3)$,and the $x$-axis.

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

The area (in square units) of the region enclosed by the two circles $x^2+y^2=1$ and $(x-1)^2+y^2=1$ is

The area of the shaded region bounded by the curves $y = \sin x$, $y = \cos x$ and the $x$-axis between $x = 2\pi$ and $x = \frac{5\pi}{2}$ is ... sq. units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module