posted by Anonymous .
Calculate the area of the surface bounded by the curves representing the functions f(x)=x²+6x and g(x)=4/(x-2).
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
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/(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
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 ...