The area bounded by the curves y = (x + 1)^2, | vedclass

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(B) The curves are $y = (x+1)^2$ and $y = (x-1)^2$. They intersect at $P(0, 1)$.The region is bounded by $y = (x+1)^2$ from $x = -1$ to $x = 0$ and $y

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$\frac{2}{3}$

Vedclass · Answer

(B) The curves are $y = (x+1)^2$ and $y = (x-1)^2$. They intersect at $P(0, 1)$.The region is bounded by $y = (x+1)^2$ from $x = -1$ to $x = 0$ and $y = (x-1)^2$ from $x = 0$ to $x = 1$.Required Area $= \int_{-1}^{0} (x+1)^2 dx + \int_{0}^{1} (x-1)^2 dx$$= \left[ \frac{(x+1)^3}{3} \right]_{-1}^{0} + \left[ \frac{(x-1)^3}{3} \right]_{0}^{1}$$= \left( \frac{1}{3} - 0 \right) + \left( 0 - (-\frac{1}{3}) \right)$$= \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

The area bounded by the curves $y = (x + 1)^2$,$y = (x - 1)^2$ and the line $y = 0$ is

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