sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle.
y = 2x + 1, y = 13 − x^2, x = −1, x = 2
To sketch the region enclosed by the given curves, we need to find the points where the curves intersect and determine the boundaries of the region.
First, let's find the points of intersection between the curves y = 2x + 1 and y = 13 - x^2:
2x + 1 = 13 - x^2
Rearranging the equation, we get:
x^2 + 2x - 12 = 0
Factoring the quadratic equation, we get:
(x + 4)(x - 3) = 0
So, x can be either -4 or 3.
Next, let's plot the given curves, y = 2x + 1 and y = 13 - x^2, as well as the vertical lines x = -1 and x = 2.
- Start by drawing the coordinate axes on a graph paper.
- Plot the point (-4, y) on the graph for the curve y = 2x + 1.
- Plot the point (3, y) on the graph for the curve y = 13 - x^2.
- Draw the two curves by connecting these plotted points on the graph.
- Draw the vertical lines x = -1 and x = 2.
Now, let's determine whether to integrate with respect to x or y.
By looking at the given curves, we can see that the region enclosed is bounded by the curves y = 2x + 1 and y = 13 - x^2 between the vertical lines x = -1 and x = 2.
Since the boundaries are defined in terms of x, we need to integrate with respect to x to find the area of the enclosed region.
Finally, to draw a typical approximating rectangle, we can choose a value of x within the interval [-1, 2], divide the interval into smaller subintervals, and draw a rectangle using the side lengths determined by the given curves at that particular x-value. Repeat this process for each subinterval, and you will have a series of approximating rectangles that help visualize the enclosed region.
since y is a function of x in both cases, it will be easier to integrate along x.
∫[-1,2] (13-x^2) - (2x+1) dx
integrating along y would require breaking the area into two parts.