The following definite integral has a known value of say 23.
$\displaystyle \int_{0}^{10}\frac{h(t)dt}{11-t}$
Without knowing h(t), is it possible to determine the value of the following:
$\displaystyle \int_{0}^{10}\frac{h(t)dt}{13-t}$
The following definite integral has a known value of say 23.
$\displaystyle \int_{0}^{10}\frac{h(t)dt}{11-t}$
Without knowing h(t), is it possible to determine the value of the following:
$\displaystyle \int_{0}^{10}\frac{h(t)dt}{13-t}$
No, it's not possible. Consider for example two examples for h(t):
1. If $\displaystyle h(t) = \frac 1 {8}(11-t)(13-t)$, then you have
$\displaystyle \int_{0}^{10}\frac{(11-t)(13-t)}{8(11-t)}dt = \int_{0}^{10} \frac{(13-t)} 8 dt = \frac 1 8 \left( 13(10)-\frac {10^2} 2 \right) = 10$
and
$\displaystyle \int_{0}^{10}\frac{(11-t)(13-t)}{8(13-t)}dt = \int_{0}^{10} \frac {(11-t)} 8 dt = \frac 1 8 \left( 11(10)-\frac {10^2} 2 \right) = \frac {15} 2$
2. Now consider $\displaystyle h(t) = \frac 1 {129} (11-t)^2(13-t)$:
$\displaystyle \small{\int_{0}^{10}\frac{(11-t)^2(13-t)}{129(11-t)}dt = \int_{0}^{10} \frac {143-24t+t^2} {129} dt =\frac 1 {129} \left( 143(10)-24(10)+10^2 \right) = 10}$
which is the same value as before. But
$\displaystyle \small{\int_{0}^{10}\frac{(11-t)^2(13-t)}{129(13-t)}dt = \int_{0}^{10} \frac {(121-22t+t^2)}{129} dt =\frac 1 {129} \left( 121(10)-22(10)+10^2 \right) = \frac {109} {129}}$
which is different than before. Hence simply knowing that $\displaystyle \small{\int_{0}^{10}\frac{h(t)}{(11-t)}dt = 10}$ is not enough information to know the value of $\displaystyle \small{\int_{0}^{10}\frac{h(t)}{(13-t)}dt}$.
Note that $\displaystyle \int_{0}^{10} \frac{h(t)dt}{13-t} = \int_{0}^{10} \left[\frac{h(t)dt}{11-t} \left(\frac{11-t}{13-t}\right)\right] = \int_{0}^{10} \left[ \frac{h(t)dt}{11-t} \left( 1 - \frac{2}{13-t} \right) \right]$
$\displaystyle = 23 - 2 \int_{0}^{10} \left(\frac{h(t)dt}{(11-t)}\frac{1}{13-t} \right)$
So you have to find a way to compute the above integral, or show that it is impossible. You could try integration by parts, but I doubt it'll work.
Thank you ebaines, Good way to explain it and prove to someone.
Thanks Richard, Are you pretty certain integration by parts won't work?
Instead of having $\displaystyle \frac{1}{11-t}$ and $\displaystyle \frac{1}{13-t}$
What if my known value was $\displaystyle \int_{0}^{10}{e^{10-t}h(t)dt}$.
Could I then calculate:
$\displaystyle \int_{0}^{10}{e^{12-t}h(t)dt}$
Thanks for the help. It's been over 20 years since I did much calculus, and I am a bit rusty.
Thank you HallsofIvy,
Last one (I hope )
What if my known value is $\displaystyle \int_0^{10}e^{-c(10-t)}h(t)dt$ and I am trying to calculate$\displaystyle \int_0^{10}e^{-c(12-t)}h(t)dt$ where c is a known constant.