Maths
posted by Anonymous .
Calculate the area of the surface bounded by the curves representing the functions f(x)=x²+6x and g(x)=4/(x2).
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 2/3ln(2) which equals 1.something

the 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 
for your first question we first have to find the intersection of the two curves.
x^2 + 6x = 4/(x2)
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/(x2)  (x^2+6x) by dx from 5.91 to .3
= 4ln(x2)  (1/3)x^3  3x^2 from ....
messy arithmetic coming up ...
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