# Thread: Calculating integral value based on similar integral

1. ## Calculating integral value based on similar integral

The following definite integral has a known value of say 23.
$\int_{0}^{10}\frac{h(t)dt}{11-t}$

Without knowing h(t), is it possible to determine the value of the following:
$\int_{0}^{10}\frac{h(t)dt}{13-t}$

2. ## Re: Calculating integral value based on similar integral

No, it's not possible. Consider for example two examples for h(t):

1. If $h(t) = \frac 1 {8}(11-t)(13-t)$, then you have

$\int_{0}^{10}\frac{(11-t)(13-t)}{8(11-t)}dt = \int_{0}^{10} \frac{(13-t)} 8 dt = \frac 1 8 \left( 13(10)-\frac {10^2} 2 \right) = 10$

and

$\int_{0}^{10}\frac{(11-t)(13-t)}{8(13-t)}dt = \int_{0}^{10} \frac {(11-t)} 8 dt = \frac 1 8 \left( 11(10)-\frac {10^2} 2 \right) = \frac {15} 2$

2. Now consider $h(t) = \frac 1 {129} (11-t)^2(13-t)$:

$\small{\int_{0}^{10}\frac{(11-t)^2(13-t)}{129(11-t)}dt = \int_{0}^{10} \frac {143-24t+t^2} {129} dt =\frac 1 {129} \left( 143(10)-24(10)+10^2 \right) = 10}$

which is the same value as before. But

$\small{\int_{0}^{10}\frac{(11-t)^2(13-t)}{129(13-t)}dt = \int_{0}^{10} \frac {(121-22t+t^2)}{129} dt =\frac 1 {129} \left( 121(10)-22(10)+10^2 \right) = \frac {109} {129}}$

which is different than before. Hence simply knowing that $\small{\int_{0}^{10}\frac{h(t)}{(11-t)}dt = 10}$ is not enough information to know the value of $\small{\int_{0}^{10}\frac{h(t)}{(13-t)}dt}$.

3. ## Re: Calculating integral value based on similar integral

Note that $\int_{0}^{10} \frac{h(t)dt}{13-t} = \int_{0}^{10} \left[\frac{h(t)dt}{11-t} \left(\frac{11-t}{13-t}\right)\right] = \int_{0}^{10} \left[ \frac{h(t)dt}{11-t} \left( 1 - \frac{2}{13-t} \right) \right]$

$= 23 - 2 \int_{0}^{10} \left(\frac{h(t)dt}{(11-t)}\frac{1}{13-t} \right)$

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.

4. ## Re: Calculating integral value based on similar integral

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 $\frac{1}{11-t}$ and $\frac{1}{13-t}$

What if my known value was $\int_{0}^{10}{e^{10-t}h(t)dt}$.

Could I then calculate:
$\int_{0}^{10}{e^{12-t}h(t)dt}$

Thanks for the help. It's been over 20 years since I did much calculus, and I am a bit rusty.

5. ## Re: Calculating integral value based on similar integral

Originally Posted by NotionCommotion
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 $\frac{1}{11-t}$ and $\frac{1}{13-t}$

What if my known value was $\int_{0}^{10}{e^{10-t}h(t)dt}$.

Could I then calculate:
$\int_{0}^{10}{e^{12-t}h(t)dt}$

Thanks for the help. It's been over 20 years since I did much calculus, and I am a bit rusty.
k
In this case you can. What you do is write $e^{12- t}= e^{2+ 10- t}= e^2e^{10-t}$ so the integral becomes $\int_0^{10}e^{12-t}h(t)dt= e^2\int_0^{10}^{10-t}h(t)dt$.

6. ## Re: Calculating integral value based on similar integral

Originally Posted by HallsofIvy
k
In this case you can. What you do is write $e^{12- t}= e^{2+ 10- t}= e^2e^{10-t}$ so the integral becomes $\int_0^{10}e^{12-t}h(t)dt= e^2\int_0^{10}^{10-t}h(t)dt$.
Cool. I miss math. Thanks!

7. ## Re: Calculating integral value based on similar integral

Thank you HallsofIvy,

Last one (I hope )

What if my known value is $\int_0^{10}e^{-c(10-t)}h(t)dt$ and I am trying to calculate $\int_0^{10}e^{-c(12-t)}h(t)dt$ where c is a known constant.

8. ## Re: Calculating integral value based on similar integral

Simplify using exponential laws, similar to how HallsofIvy did three posts earlier...

9. ## Re: Calculating integral value based on similar integral

Thank you Prove It. I actually vaguely remember the laws, but not perfectly. After longer than I expected googleing, I confirmed that e^ab=(e^a)^b. Great, but where do I go from there? Thanks!

10. ## Re: Calculating integral value based on similar integral

Never mind. Answer was simply $e^{-2c}$ Thank you all again for your help!

11. ## Re: Calculating integral value based on similar integral

Originally Posted by NotionCommotion
Never mind. Answer was simply $e^{-2c}$ Thank you all again for your help!
Assuming you mean \displaystyle \begin{align*} e^{-2c} \end{align*} times the original integral, then you would be correct.