Finding area of curve....

Dec 2009
55
0
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. (Do this on paper. Your instructor may ask you to turn in this graph.)


This is the solution I got, but it was marked wrong:

x^2-2x = x+4

x^2-2x-x-4 =0
x^2-3x-4 =0
(x+1)(x-4), bounds are (-1,4)
= (int from -1 and 4) x^2-3x-4
=(x^3/3)-(3x^2/(2))-4x
when I plug it in, I get 191/6, but it's wrong, don't know what I did wrong..please help!
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
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. (Do this on paper. Your instructor may ask you to turn in this graph.)


This is the solution I got, but it was marked wrong:

x^2-2x = x+4

x^2-2x-x-4 =0
x^2-3x-4 =0
(x+1)(x-4), bounds are (-1,4)
= (int from -1 and 4) x^2-3x-4

your integrand is incorrect ... should be top function - bottom function.
\(\displaystyle \int_{-1}^4 (x+4) - (x^2-2x) \, dx\)
 
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