How can you evaluate this integral using an analytic method?

[JulieK]

I know the value of the following definite integral

\int_{a}^{b}ydx

I also have a realtion

x=f(y)

i.e. x is an explicit function of y but I do not have y as an explicit
function of x. The relation between x and y is generally non linear.

Now I want to get the following definite integral

\int_{a}^{b}\left[\int ydx\right]xdx

i.e. \int ydx multiplied by x evaluated over the interval [a,b].

Is there an analytic (not numeric) way to evaluate this integral using
for example mean value or similar averaging technique?

 

[acegikmoqsuwy]

Well, I think all you need to do is to find dx in terms of dy and then it should be simple since you already have x in terms of y, just sub in the integral.

dx=f'(y)dy
\displaystyle\large\therefore \int_a^b \left[\int ydx\right]xdx=\int_a^b \left[\int yf'(y)dy\right]f(y)f'(y)dy

You already have the value of the integral \displaystyle\large\int_a^b ydx=\int_a^b yf'(y)dy


I believe (but do not recall) that there is a way to use sort of the "opposite" of the mean value theorem.

Also, if possible, it would simply be easy enough to take f^{-1}(y)

 

[I like Serena]

I know the value of the following definite integral

\int_{a}^{b}ydx

I also have a realtion

x=f(y)

i.e. x is an explicit function of y but I do not have y as an explicit
function of x. The relation between x and y is generally non linear.

Now I want to get the following definite integral

\int_{a}^{b}\left[\int ydx\right]xdx

i.e. \int ydx multiplied by x evaluated over the interval [a,b].

Is there an analytic (not numeric) way to evaluate this integral using
for example mean value or similar averaging technique?

Your integral is actually:
\begin{aligned}\int_{a}^{b}\left[\int ydx\right]xdx
&=\int_{a}^{b}\left[\int y(\xi) d\xi\right]xdx \\
&= \int_{a}^{b} xdx \cdot \int y(\xi) d\xi \\
&= \frac 1 2(b^2-a^2)\int y(\xi) d\xi \\
&= \frac 1 2(b^2-a^2) \int y dx \\
\end{aligned}