On-line Math 21

6.1  Area between curves

Example 2 Find the areas enclosed by the curves y = arctan(x) and y = px/4 . The constants are supposed to help.

Solution

The region is as pictured:

Ah, but how do you do that integral?

Here is the answer:

ó õ arctan(x)dx = x·arctan(x)- 1 2 ln| 1+x2| +C.

OK, the absolute value signs are not important. Then plug in to the definite integral,

Area = 2 ó õ 1 0  æ ç è arctan(x)- px 4 ö ÷ ø dx = 2 æ ç è x·arctan(x)- 1 2 ln| 1+x2| - px2 8 ö ÷ ø ê ê ê 1 0  = 2 æ ç è 1·arctan(1)- 1 2 ln| 1+12| - p12 8 ö ÷ ø - æ ç è 0·arctan(0)- 1 2 ln| 1+02| - p02 8 ö ÷ ø = 2 æ ç è 1 p 4 - 1 2 ln(2)- p 8 ö ÷ ø = p 4 -ln(2) @ 0.092

Remark By the way, when I first solved this, I dropped that factor of 1/2 in front of the natural log, resulting in an answer that was negative. I only noticed that when I computed the numerical approximation (with a computer, of course). My reaction was what yours should be, look back in the computation to see why I got a negative area. These ``applied'' problems provide a reality check; since areas are always positive (except in Star Trek ), if you get a negative area, something was wrong.

Copyright (c) 2000 by David L. Johnson.