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Chapter 12: Problem 1088

The area enclosed by the curves \(x^{2}=y, y=x+2\) and \(x\) -axis is...... (a) \((3 / 2)\) (b) \((5 / 2)\) (c) \((5 / 6)\) (d) \((7 / 6)\)

Short Answer

Expert verified
The area enclosed by the curves \(x^2 = y, y = x + 2\) and the x-axis is (b) \(\frac{5}{2}\).

Step by step solution

01

Find the Intersection Points

Let's first find the x-coordinates of the points where the two curves intersect. To do that, we set the two equations equal to each other: \(x^2 = x + 2\) Now, we can rearrange the equation to make it a quadratic equation: \(x^2 - x - 2 = 0\) Now, let's solve the quadratic equation for x.
02

Solve the Quadratic Equation

Using the quadratic formula, we can solve the equation \(x^2 - x - 2 = 0\). The quadratic formula is given by: \(x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\) In our case, a = 1, b = -1, and c = -2. So, \(x = \frac{-( -1)\pm\sqrt{( -1)^2-4(1)( -2)}}{2(1)}\) Calculating, we get x = -1 and x = 2 as the x-coordinates of the intersection points.
03

Find the y-coordinates of Intersection Points

Now, let's find the y-coordinates of the intersection points by substituting the x-coordinates into either of our given equations (let's use the first one: x^2 = y): For x = -1: \(y = (-1)^2 = 1\) For x = 2: \(y = (2)^2 = 4\) So, the intersection points are (-1, 1) and (2, 4).
04

Set up the Area Integral

Now we can set up the integral, representing the area enclosed by the curves, between -1 and 2: \(A = \int_{-1}^2 [(x+2) - x^2] dx\)
05

Evaluate the Integral

Now, let's evaluate the integral: \(A = \int_{-1}^2 (x+2 - x^2) dx\) \(A = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^2\) Now, let's find the value of the expression at the bounds and subtract: \((\frac{(2)^2}{2}+2(2)-\frac{(2)^3}{3}) - (\frac{(-1)^2}{2}+2(-1)-\frac{(-1)^3}{3})\) After evaluating the expression, we get: \(A = \frac{5}{2}\) So, the answer is (b) \(\frac{5}{2}\).

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