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July 28, 2014

July 28, 2014

Posted by **Anonymous** on Wednesday, March 10, 2010 at 8:01am.

Thank you in advance

If possible, help with this question aswell: Calculate f'(x)=(e^2x - ln(2)). Between ln(2) and 0.

I know you integrate the function and then put in the values. But I don't seem to be getting the correct answer which is apparently 3/2 - ln(2), which equals 0.807

I get 2/3-ln(2) which equals 1.something

- Maths -
**Reiny**, Wednesday, March 10, 2010 at 8:30amthe integral of e^2x - ln(2) is

(1/2)e^(2x) - (ln2)x from 0 to ln2

= (1/2)(4) - (ln2)^2 - ((1/2((1) - 0)

= 2 - 1/1 - (ln2)^2

= 3/2 - (ln2)^2

- Maths -
**Reiny**, Wednesday, March 10, 2010 at 8:40amfor your first question we first have to find the intersection of the two curves.

x^2 + 6x = 4/(x-2)

x^3 + 6x^2 - 12x - 4 = 0

I could not find any rational roots so I used my "trusty" cubic equation solver at

http://www.1728.com/cubic.htm

and got x = 2.22, -5.91, and -.3

making a sketch shows the only closed region is between -5.91 and -.3

So the area is

Integral [4/(x-2) - (x^2+6x) by dx from -5.91 to -.3

= 4ln(x-2) - (1/3)x^3 - 3x^2 from ....

messy arithmetic coming up ...

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