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

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(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

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

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(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

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