Integrals of differentialble functions on a bounded interval

[mjpam]

Given two functions f_{1}(x) and f_{2}(x) that are differentiable on [a_{1},a_{2}] and \int^{a_2}_{a_1}f_1(x)f'_{2}(x)dx=b_2, how would one calculate \int^{a_2}_{a_1}f_1(x)f'_{2}(x)dx?

This is not a homework problem. I saw it on the internet and realized that I did not know where to begin solving it.

 

[HallsofIvy]

I am confused. Are you asking how to find b_2? If so, that would depend strongly on what f_1 and f_2 are. There are many different ways to evaluate various integrals, and, in fact, if we take f_x(x)= 1, for all x, and f_2(x)= e^{-x^2}, both differentiable functions, then it is easy to show that
\int_{a_1}^{a_2}f_1(x)f_2(x)dx= \int_{a_1}^{a_2}e^{-x^2}dx
exists but there is no analytic method of finding the integral. The best you could do is use some numeric method, such as Simpson's rule, to find the value.

 

[micromass]

Are you perhaps asking how to calculate \int_{a_1}^{a_2}{f_1^\prime(x)f_2(x)dx}?? This can be easily done by "integration by parts".

The answer is \int_{a_1}^{a_2}{f_1^\prime(x)f_2(x)dx}=f_1(a_2)f_2(a_2)-f_1(a_1)f_1(a_2)-b_2.

 

[mjpam]

Are you perhaps asking how to calculate \int_{a_1}^{a_2}{f_1^\prime(x)f_2(x)dx}?? This can be easily done by "integration by parts".

The answer is \int_{a_1}^{a_2}{f_1^\prime(x)f_2(x)dx}=f_1(a_2)f_2(a_2)-f_1(a_1)f_1(a_2)-b_2.

I think that's correct.

My question was:

Given two functions f_{1}(x) and f_{2}(x) differentiable on a closed interval [a_{1},a_{2}], two constants c_{1} and c_{2}, and the facts that \int_{a_{1}}^{a_{2}}f_{1}f_{2}^{\prime}dx=c_{1} and \int_{a_{1}}^{a_{2}}f_{1}^{\prime}f_{2}dx=c_{2}, can c_{2} be expressed in terms of c_{1}?