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
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 ...
To calculate the area of the surface bounded by the curves representing the functions f(x) = x^2 + 6x and g(x) = 4/(x - 2), you need to find the points of intersection first.
To do this, set the two functions equal to each other:
x^2 + 6x = 4/(x - 2)
Multiply both sides by (x - 2) to eliminate the fraction:
(x^2 + 6x)(x - 2) = 4
Expand and rearrange the equation:
x^3 - 2x^2 + 6x^2 - 12x - 4 = 0
Combine like terms:
x^3 + 4x^2 - 12x - 4 = 0
Now, to solve this equation and find the values of x where the curves intersect, you can use numerical methods such as graphing or a calculator.
Once you have the x-coordinate values of the points of intersection, you can calculate the area between the curves using integration.
For example, if you find that the curves intersect at x = a and x = b, where a < b, then the area can be calculated as follows:
Area = ∫[a, b] (f(x) - g(x)) dx
In this case, the equation of the area would be:
Area = ∫[a, b] [(x^2 + 6x) - (4/(x - 2))] dx
To accurately calculate the definite integral, you can use various integration techniques such as the power rule, substitution method, or partial fractions. Once you find the antiderivative, evaluate it from a to b and compute the difference to find the desired area.
Regarding the second question about calculating f'(x) = e^(2x) - ln(2) between ln(2) and 0, you mentioned that you obtained 2/3 - ln(2), which is incorrect. Let's go through the correct solution:
To find the derivative of f(x) = e^(2x) - ln(2), you need to take the derivative term by term. The derivative of e^(2x) is 2e^(2x) using the chain rule, and the derivative of ln(2) is 0 since it is a constant.
So, f'(x) = 2e^(2x) - 0 = 2e^(2x)
To evaluate f'(x) between ln(2) and 0, substitute the upper and lower limits into the derivative expression:
f'(ln(2)) = 2e^(2ln(2))
f'(0) = 2e^(2(0))
Simplifying these expressions:
f'(ln(2)) = 2e^(ln(2))^2 = 2e^(ln(2 * 2)) = 2e^ln(4) = 2 * 4 = 8
f'(0) = 2e^0 = 2 * 1 = 2
Therefore, the difference between f'(ln(2)) and f'(0) is:
f'(ln(2)) - f'(0) = 8 - 2 = 6
So, the correct answer to the second question is 6, not 0.807.
I hope this helps! Let me know if you have any further questions.