How do you find the area of the region bounded above by the parabola y=2-x^2, and below by the line y=-x' ?

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How do you find the area of the region bounded above by the parabola y=2-x^2, and below by the line y=-x' ?

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Answer 1 · 2018-05-14T15:28:43

Use the form:

$Area = \int_{a}^{b} f_{2} (x) - f_{1} (x) d x, f_{2} (x) > f_{1} (x)$ $"Area" = int_a^b f_2(x)-f_1(x) dx, f_2(x) > f_1(x)$

Explanation:

We know that $f_{2} (x) = 2 - x^{2}$ $f_2(x) = 2-x^2$ and $f_{1} (x) = - x$ $f_1(x) = -x$

$Area = \int_{a}^{b} 2 - x^{2} - (- x) d x$ $"Area" = int_a^b 2-x^2-(-x) dx$

Find the values of $a$ $a$ and $b$ $b$ by setting the right sides of the two equations equal:

$- x = 2 - x^{2}$ $-x = 2-x^2$

$x^{2} - x - 2 = 0$ $x^2-x-2=0$

$(x - 2) (x + 1) = 0$ $(x-2)(x+1)=0$

$x = - 1$ $x = -1$ and $x = 2$ $x = 2$

This means that $a = - 1$ $a = -1$ and $b = 2$ $b = 2$

$Area = \int_{- 1}^{2} 2 - x^{2} + x d x$ $"Area" = int_-1^2 2-x^2+x dx$

$Area = 2 x - \frac{1}{3} x^{3} + \frac{1}{2} x^{2} {∣ ∣ ∣}_{- 1}^{2}$ $"Area" = 2x-1/3x^3+1/2x^2|_-1^2$

$Area = 2 (2) - \frac{1}{3} {(2)}^{3} + \frac{1}{2} {(2)}^{2} - (2 (- 1) - \frac{1}{3} {(- 1)}^{3} + \frac{1}{2} {(- 1)}^{2})$ $"Area" = 2(2)-1/3(2)^3+1/2(2)^2-(2(-1)-1/3(-1)^3+1/2(-1)^2)$

$Area = \frac{9}{2}$ $"Area" = 9/2$

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How do you find the area of the region bounded above by the parabola y=2-x^2, and below by the line y=-x' ?

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