Question

In Exercises \(15-34,\) find the area of the regions enclosed by the lines and curves. $$x=y^{2} \quad$$ and $$\quad x=y+2$$

Short answer

The area of the region enclosed by the curves is \(4.5\) square units.

Step-by-step solution

Sketch the curves

Initially, graph the functions \(x = y^2\) and \(x = y+2\) to understand the region enclosed by them. Graphing these functions, it can be observed that \(x = y^2\) is a parabola and \(x = y+2\) is a straight line.

Find Points of Intersection

Set the two equations equal to each other and solve for \(y\) to find the points where the graphs intersect. Solve \(y^2 = y+2\). This gives the solutions \(y=-1,2\) which correspond to points (-1, -1) and (4, 2). These are our limits of integration.

Setup the Integral to Find Area

The area between the two curves is the definite integral from the lower intersection point to the upper intersection point of the absolute value of the difference between the two functions. Here, the area \(A\) is given by \(A = - \int_{-1}^{2} y^2 - (y + 2) dy\).

Evaluate the Integral

Now, the integral can be evaluated using the Fundamental Theorem of Calculus. This gives \(A = - \left[\frac{y^3}{3} - \frac{y^2}{2} - 2y \right]_{-1}^{2}\) which simplifies to \(A=9/2\) or \(4.5\) square units.