# Thread: Substituting numbers into an integral.

1. ## Substituting numbers into an integral.

My problem is that I can't seem to substitute numbers into an integral properly !

One of my concerns is that my evaluation of I0 is different if I use the integral or use the expression (-sinh1 and sinh1 respectively). Since the expression is correct, as it was a "prove that" question and therefore written in the book, i'm not really sure why i'm getting two different answers.

I can't see any problem with my working for part (b). That's just one of the reasons why this is so infuriating!

Anyway, if anyone could see what i'm doing wrong it would be a great help.

2. I managed to find another question that I can't do with the same problem: I can't substitute numbers into an expression!

3. Hi,

For the first problem, question a), I don't see why you say it's wrong... ?

Now, for question b), I know there's a problem :

$\int_1^0 x \cosh(x) ~dx=[x \sinh(x)]_1^0-\int_1^0 \sinh(x) ~dx$

But $[x \sinh(x)]_1^0=[f(x)]_1^0=f(0)-f(1)={\color{red}-} \sinh(1)$

And $\int_1^0 \sinh(x) ~dx=\cosh(0)-\cosh(1)=1-\cosh(1)$

$\implies I_1={\color{red}-} \sinh(1)+\cosh(1)-1$

Now, there's a possibility that your answers have been given in a form with exponentials.

Just use :
$\cosh(x)=\frac{e^x+e^{-x}}{2}$ and $\sinh(x)=\frac{e^x-e^{-x}}{2}$

4. Originally Posted by Showcase_22
I managed to find another question that I can't do with the same problem: I can't substitute numbers into an expression!

It's normal

$I_{5,6}=\frac{5!6!}{11!} \cdot {\color{red}I_{0,11}}$

5. For part a) my book gives the answer as 37sinh1-28cosh1. They've got this answer by making I0=sinh1 instead of -sinh1. I just thought I needed a second opinion before I branded the book incorrect.

For the second question they get:

I couldn't see where my answer was going wrong because it looks so straightforward!

Thanks for pointing out my mistake for the second question. The boundaries for the integral are the other way around from what i'm used to.

6. Originally Posted by Showcase_22
For part a) my book gives the answer as 37sinh1-28cosh1. They've got this answer by making I0=sinh1 instead of -sinh1. I just thought I needed a second opinion before I branded the book incorrect.
Well, I've check several times... all getting -sinh(1)
I don't see where the mistake would be

For the second question they get:

I couldn't see where my answer was going wrong because it looks so straightforward!
Yes, if you calculate I_(0,11), you'll get a factor 2^(12)

Thanks for pointing out my mistake for the second question. The boundaries for the integral are the other way around from what i'm used to.
Ya I know... If it really disturbs you, you can work on $\int_0^1 \dots ~dx=-\int_1^0 \dots ~dx$

7. hmmm.

The general consensus is that a) is right but incorrect in the book and b) was incorrectly done by me (whoops!).

I finally get their answer for the second question! I didn't include I_(0,11) because I thought it was 0 since I used the expression (darn!!). Anyhow, it's great that I finally understand what i'm doing!

8. Originally Posted by Showcase_22
hmmm.

The general consensus is that a) is right but incorrect in the book and b) was incorrectly done by me (whoops!).
Okay, I've figured out the problem...

The formula is :

$\boxed{I_n={\color{red}-\sinh(1)+n \cosh(1)}+n(n-1) I_{n-2}}$

I thought the formula was ok because it was the book's ^^'

I finally get their answer for the second question! I didn't include I_(0,11) because I thought it was 0 since I used the expression (darn!!). Anyhow, it's great that I finally understand what i'm doing!

Yep, that was the missing step ^^ I thought you already knew when you last posted !

9. lol, I know i'm a little slow. In my weak (but flimsy!) defence I was stuck on some other questions as well.

Since I now regard this place as the fountain of all mathematical knowledge, I might post them since i've just about reached the end of my tether!!

Thanks for your help! You're pretty darn good at this maths malarkey!