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Chapter 8: Problem 9

Find the area of the region bounded on the right by the line given by \(x+y=2\), on the left by the parabola given by \(y=x^{2}\), and below by the \(x\) -axis.

Short Answer

Expert verified
The area of the region bounded by the line \(x+y=2\), the parabola \(y=x^2\), and the x-axis is \(\frac{3}{2}\) square units.

Step by step solution

01

Find Points of Intersection

To find the points of intersection between the line and the parabola, we can set them equal to each other: \(x + y = 2\) \(y = x^2\) Substitute the second equation into the first: \(x + x^2 = 2\) Rearrange to find a quadratic equation: \(x^2 + x - 2 = 0\) Factor the quadratic equation: \((x + 2)(x - 1) = 0\) Solve for x: \(x = -2, x = 1\) Now, we can plug these x-values back into one of the original equations to find the corresponding y-values. We'll use the first equation: For x = -2: \(y = 2 - (-2) = 4\) For x = 1: \(y = 2 - 1 = 1\) So the points of intersection are \((-2, 4)\) and \((1, 1)\).
02

Set up the integral

To find the area of the region bounded by the graphs, we need to subtract the function that represents the parabola from the function that represents the line. Then, we integrate the resulting function over the interval \([-2, 1]\): Area = \(\int_{-2}^{1} (x + y - x^2) dx\) Since we know y = x^2, we can rewrite the integral as: Area = \(\int_{-2}^{1} (x + x^2 - x^2) dx\) Simplify the integrand: Area = \(\int_{-2}^{1} x dx\)
03

Evaluate the integral

We can find the antiderivative of the integrand and then evaluate it at the bounds of the interval to find the area: The antiderivative of x is \(\frac{x^2}{2}\): Area = \(\frac{1^2}{2} - \frac{(-2)^2}{2}\) Area = \(\frac{1}{2} - \frac{4}{2}\) Area = \(-\frac{3}{2}\) Since the area can't be negative, we'll take the absolute value of the result: Area = \(\frac{3}{2}\) The area of the region bounded by the given line and parabola and the x-axis is \(\frac{3}{2}\) square units.

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