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

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(D) To find the area bounded by the curve $y = x^2 - 1$ and the line $y = 3 - x$,we first find the points of intersection by setting the equations equ

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$\frac{9}{2}$

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(D) To find the area bounded by the curve $y = x^2 - 1$ and the line $y = 3 - x$,we first find the points of intersection by setting the equations equal: $x^2 - 1 = 3 - x$ $x^2 + x - 4 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-4)}}{2(1)} = \frac{-1 \pm \sqrt{17}}{2}$ Let $x_1 = \frac{-1 - \sqrt{17}}{2}$ and $x_2 = \frac{-1 + \sqrt{17}}{2}$. The area $A$ is given by: $A = \int_{x_1}^{x_2} [(3 - x) - (x^2 - 1)] dx = \int_{x_1}^{x_2} (4 - x - x^2) dx$ $A = [4x - \frac{x^2}{2} - \frac{x^3}{3}]_{x_1}^{x_2}$ Using the property that for $ax^2 + bx + c = 0$ with roots $\alpha, \beta$,the integral $\int_{\alpha}^{\beta} (ax^2 + bx + c) dx = -\frac{a}{6}(\beta - \alpha)^3$,where $a = -1$ and $(\beta - \alpha) = \sqrt{17}$: $A = -\frac{-1}{6} (\sqrt{17})^3 = \frac{17\sqrt{17}}{6}$.

The area bounded by the curve $y = x^2 - 1$ and the straight line $x + y = 3$ is:

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