Double Integrals and Utilizing the Mean Value Theorem.

[ardentmed]

Hey guys,

I'm having trouble with this problem set I'm working on at the moment. I'd appreciate some help with this question:

Find f(t) if f"(t) = 2e^(-t) + cos(pi*t), f(0) = f(1) = 0. Give exact answers.

Alright, so I simply integrated f''(x) up to f(t) given f(0) = 0 and f(1) = 0. This necessitates the mean value theorem to find f'(c), which is a coordinate for f'(t).

Therefore, using the theorem, I computed f'(0) = -2 + C, where C=2 (computed via substitution).

Also, I took the integral of f'(t) to compute the final answer, which is:

f(t) = 2e^(-t) - (cosπt/ π^2) + 2t + 1/(π^2) - 2.

Am I on the right track?

Thanks in advance, guys.

Thanks in advance.

 

[Ishuda]

I think you are making the problem too complicated. If you do the double integration you get

f(t) = 2 e^(-t) - (cosπt/ π^2) + a t + b

where a and b are constants to be determined. As you indicated, f(0) = 0 gives b = 1/(π^2) - 2. Now
f(1) = 2/e + 1 / π^2 + a + b.
Since f(1) is zero we have
a = - (2/e + 1 / π^2 + b) = -2 (1/e + 1 / π^2 - 1)

 

[ardentmed]

I think you are making the problem too complicated. If you do the double integration you get

f(t) = 2 e^(-t) - (cosπt/ π^2) + a t + b

where a and b are constants to be determined. As you indicated, f(0) = 0 gives b = 1/(π^2) - 2. Now
f(1) = 2/e + 1 / π^2 + a + b.
Since f(1) is zero we have
a = - (2/e + 1 / π^2 + b) = -2 (1/e + 1 / π^2 - 1)

Alright, so how would I go about finding "b" with your method?


Thanks for the help.

I got b= 1/pi^2 -2 after substituting f(0) into the equation.

Also, I ultimately ended up getting.

f(t) = 2 e^(-t) - (cosπt/ π^2) -2t((1/e + 1 / π^2 - 1)) + 1/pi^2 -2