Link to book index:

http://www.thescienceforum.com/Calcu...ions-8994t.php

Chapter 6: Applications of Integration

5) Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. Then find the area of the region.

y = x + 1

y = 9 - x<sup>2</sup>

x = -1

x = 2

My work:

I'll skip image posting and just describe the graph.

y = x + 1 has a slope of 1, a y-intercept of 1, and an x-intercept of -1.

y = 9 - x<sup>2</sup> is a downward turned parabola with a critical point at (0,9), a y-intercept of 9, and two x-intercepts of 3 and -3.

These two lines intersect at -3.37 and 2.37, and if we integrate with respect to x and use -1 < x < 2 (or equal to of course) to restrict the x values of our area, then these points of intersection are outside of our restricted area and of no consequence. We can then set up our integral as

∫<sub>-1</sub><sup>2</sup> [ (9-x<sup>2</sup>) - ( x + 1 ) ] dx

or

∫<sub>-1</sub><sup>2</sup> ( -x<sup>2</sup> - x + 10 ) dx

[-x<sup>3</sup>/3 - x<sup>2</sup>/2 + 10x]<sub>-1</sub><sup>2</sup>

(-8/3 - 2 + 20 ) - (1/3 - 1/2 -10 )

-3 - 3/2 + 30

51/2 or 25.5

The book answer is 19.5