The area of the region bounded by the parabola y = x2 + 1 and the straight line x + y = 3 is given by

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KollegeApply · Accepted Answer

$\frac{9}{2}$ To find the point of intersection of the parabola y = x2 + 1 and the line x + y = 3 substitute y = 3 − x in y = x2 + 1 3 - x = x^2 + 1 \Rightarrow x^2 + x - 2 = 0 \Rightarrow \left( x - 1 \right)\left( x + 2 \right) = 0 \Rightarrow x = 1\text{ or }x = - 2 \therefore y = 2\text{ or }y = 5 So, we get the points of intersection A(−2, 5) and C(1, 2). Therefore, the required area ABC, A = \int_{- 2}^1 \left( y_1 - y_2 \right) dx ............\left(\text{Where, }y_1 = 3 - x\text{ and }y_2 = x^2 + 1 \right) = \int_{- 2}^1 \left[ \left( 3 - x \right) - \left( x^2 + 1 \right) \right] d x = \int_{- 2}^1 \left( 3 - x - x^2 - 1 \right) d x = \int_{- 2}^1 \left( 2 - x - x^2 \right) d x = \left[ 2x - x^$\frac{2}{2}$ - x^$\frac{3}{3}$ \right]_{- 2}^1 = \left[ 2\left( 1 \right) - \frac{\left( 1 \right)^2}{2} - \frac{\left( 1 \right)^3}{3} \right] - \left[ 2\left( - 2 \right) - \frac{\left( - 2 \right)^2}{2} - \frac{\left( - 2 \right)^3}{3} \right] = \left[ 2 - $\frac{1}{2}$ - $\frac{1}{3}$ \right] - \left[ - 4 - 2 + $\frac{8}{3}$ \right] = 2 - $\frac{1}{2}$ - $\frac{1}{3}$ + 4 + 2 - $\frac{8}{3}$ = 8 - $\frac{1}{2}$ - $\frac{9}{3}$ = 5 - $\frac{1}{2}$ = $\frac{9}{2}$\text{ square units } [Solution diagram: 44034_solution_1.png]

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