# Integration -- basic problems

• June 12th 2009, 12:55 PM
sinewave85
Integration -- basic problems
I don't think I really understand what I am doing. This topic has me a little befuddled, so please bear with me.

First, how do you correctly express the integration of a constant? (And no, my text does not explain this. I have looked several times.)

As in:

$\int_{-3}^{-1}5dx$

I understand that the answer is 10 (the area bound by the lines $y=5, y=0, x=-1$ and $x=-3$) but not how to obtain or express that answer using the definite integral.

$\int_{a}^{b}f(x)dx=\lim_{\|P\|\to0}\sum_{k=1}^{n}f (x_{k}^{*})\Delta x$

Does not seem to make sense when f(x) is a constant.

Thanks for the help! Integration may be my waterloo.
• June 12th 2009, 01:08 PM
craig
Quote:

Originally Posted by sinewave85
I don't think I really understand what I am doing. This topic has me a little befuddled, so please bear with me.

First, how do you correctly express the integration of a constant? (And no, my text does not explain this. I have looked several times.)

As in:

$\int_{-3}^{-1}5dx$

If you integrate any constant, lets say $n$, then $\int n dx = nx$, $\int n dt = nt$ etc...

Does this help?
• June 12th 2009, 01:09 PM
Ruun
In a definite integral (i.e., with limits for the variable(s) that you're integrating with respect of), there are no integration constants.

Example:

$\int_{-1}^{-3}5dx=\frac{5}{2}x^2$ and now you evaluate the function, in virtue of Barrow's Rule: $5\left(\frac{-1^2}{2}-\frac{-3^2}{2}\right)$

In the other hand, if your integral is $\int 5dx=\frac{5}{2}x^2 + C_{0}$ where $C_{0}=f(t_{0})$ for some $t_0$ in the domain of the function.

EDIT: Ok, bad english.I'm sorry, I didn't looked as well as I should to your question (Worried)
• June 12th 2009, 01:11 PM
craig
Quote:

Originally Posted by sinewave85
but not how to obtain or express that answer using the definite integral.

When you have integrated you constant, you solve it like you would any normal definite integral, by putting in the limits and getting your value.
• June 12th 2009, 01:12 PM
craig
Quote:

Originally Posted by Ruun
In a definite integral (i.e., with limits for the variable(s) that you're integrating with respect of), there are no integration constants.

Example:

$\int_{-1}^{-3}5dx=\frac{5}{2}x^2$ and now you evaluate the function, in virtue of Barrow's Rule: $5\left(\frac{-1^2}{2}-\frac{-3^2}{2}\right)$

In the other hand, if your integral is $\int 5dx=\frac{5}{2}x^2 + C_{0}$ where $C_{0}=f(t_{0})$ for some $t_0$ in the domain of the function.

This is wrong, $\int 5dx$ does not equal $\frac{5}{2}x^2 + C_{0}$
• June 12th 2009, 01:14 PM
sinewave85
Quote:

Originally Posted by craig
If you integrate any constant, lets say $n$, then $\int n dx = nx$, $\int n dt = nt$ etc...

Does this help?

Where $x \mbox{ or } t = \Delta x \mbox{ or } \Delta t$ (the distance from a to b on the definite integral $\int_{a}^{b}$)?
• June 12th 2009, 01:23 PM
craig
Quote:

Originally Posted by sinewave85
Where $x \mbox{ or } t = \Delta x \mbox{ or } \Delta t$ (the distance from a to b on the definite integral $\int_{a}^{b}$)?

Let me give you an example.

Let's say we have $\int_{0}^{5} 3 dx$, this gives us $\left[3x \right]^5_{0}$.

Expanding this we get $3(5) - 3(0) = 15$.

You could also double check this by drawing the graph of $y = 3x$ and finding the area between 0 and 5, this would give you a rectangle with sides of 3 and 5, giving you the area 15.

Does this explain it a bit more?
• June 12th 2009, 01:28 PM
sinewave85
Quote:

Originally Posted by craig
Let me give you an example.

Let's say we have $\int_{0}^{5} 3 dx$, this gives us $\left[3x \right]^5_{0}$.

Expanding this we get $3(5) - 3(0) = 15$.

You could also double check this by drawing the graph of $y = 3x$ and finding the area between 0 and 5, this would give you a rectangle with sides of 3 and 5, giving you the area 15.

Does this explain it a bit more?

Yes, that helps, thanks! I was making it much harder than it really was. I am not so sure that I have really grasped the notation for all of this, so I think that I had better go back and review before I go much further. Thanks again for the clarification.
• June 12th 2009, 01:30 PM
sinewave85
Quote:

Originally Posted by craig
When you have integrated you constant, you solve it like you would any normal definite integral, by putting in the limits and getting your value.

Is this the fundamental theorum?
• June 12th 2009, 01:32 PM
craig
Quote:

Originally Posted by sinewave85
Yes, that helps, thanks! I was making it much harder than it really was. I am not so sure that I have really grasped the notation for all of this, so I think that I had better go back and review before I go much further. Thanks again for the clarification.

It possible that we use different notation, I had never come across the $\Delta$ symbol before.

In mine above, the first one is the integral of 3 with respect to x, between x equals 5 and 0.

The second with the square brackets is when I have integrated the equation, and the limits upper and lower limits are placed at the top and bottom of the last square brackets respectively.

This is how I have been taught them anyway, if you have been taught different the continue to use the way that feels best for you.
• June 12th 2009, 01:34 PM
craig
Quote:

Originally Posted by sinewave85
Is this the fundamental theorum?

Afraid I cannot answer this, not come across the fundamental theorem before.

If it helps think of the number 5 as $5x^0$, the integrate using your usual rules.
• June 12th 2009, 01:39 PM
sinewave85
Quote:

Originally Posted by craig
Afraid I cannot answer this, not come across the fundamental theorem before.

If it helps think of the number 5 as $5x^0$, the integrate using your usual rules.

Flipping ahead a few lessons in my book I found "The First Fundamental Theorem of Calculus" (very portentious, huh?) which basicly states that the definite integral can be computed the way you show. I think that since that is further along than I am, it is only expected that I compute it as an area -- but your way is much easier. Thanks again for all of the help!
• June 12th 2009, 01:42 PM
craig
Quote:

Originally Posted by sinewave85
Flipping ahead a few lessons in my book I found "The First Fundamental Theorem of Calculus" (very portentious, huh?) which basicly states that the definite integral can be computed the way you show. I think that since that is further along than I am, it is only expected that I compute it as an area -- but your way is much easier. Thanks again for all of the help!

No problem, glad my explanations weren't too confusing ;)

Edit - just read up on the "The First Fundamental Theorem of Calculus" (never actually got taught this name), and yes that's what I've shown you above.