# Math Help - Change in integral begin/end values

1. ## Change in integral begin/end values

I am just starting to learn about integrals. We are using dummy variables to solve integrals from say 1 to 3. When we finish the range of the integral changes and I am at a loss as to how to determine the new values.

For example if I have Integral from 0 to 1 of x (2 + 3x^2) dx

it becomes integral from 2 to 5 of u^1/2 du times 1/6

PLease explain how the 0 to 1 becomes 2 to 5 and since u = 2 + 3x^2 what happened to the x in the integral that precedes the (2 + 3x^2) ^1/2?????

Either my book is missing some basic explanations or I am just slowwwwww!

Also how do I write an integral sign on this site????

2. Hello,

[math ]\int_0^1[/tex]

When doing a substitution you also have to change the values of the integral (0 and 1).

I'll make the steps...

$\int_0^1 x(2+3x^2) dx$

$u=2+3x^2$

So if $x=0$, $u=2+3*0^2=2$
If $x=1$, $u=2+3*1^2=5$

This explains the changes 0 and 1 to 2 and 5.

Plus, you have to "change" dx, and express it in respect to u.

$\frac{du}{dx}=6x$

Hence $dx=\frac{du}{6x}$

The integral becomes :

$\int_2^5 {\color{red} x} u \frac{du}{6 {\color{red} x}} = \int_2^5 \frac{u}{6} du$

3. Originally Posted by Frostking
I am just starting to learn about integrals. We are using dummy variables to solve integrals from say 1 to 3. When we finish the range of the integral changes and I am at a loss as to how to determine the new values.

For example if I have Integral from 0 to 1 of x (2 + 3x^2) dx

it becomes integral from 2 to 5 of u^1/2 du times 1/6

PLease explain how the 0 to 1 becomes 2 to 5 and since u = 2 + 3x^2 what happened to the x in the integral that precedes the (2 + 3x^2) ^1/2?????

Either my book is missing some basic explanations or I am just slowwwwww!
If $u = 2 + 3x^2$, then what was the beginning limit of $x=0$ becomes $u=2+3(0)^2 \rightarrow 2$, and the upper limit of $x=1$ becomes $u=2+3(1)^2 \rightarrow 5$.

Personally, I never change the limits as I always re-sub for x back into the integrated equation. This takes away from the annoying process of re-evaluating the limits of integration.

Originally Posted by Frostking
Also how do I write an integral sign on this site????
Type the [math ] [/tex] tags around \int like so: $\int_0^1 \sqrt{2+3x^2}dx$

4. Oh yeah, I think you forgot ^1/2 in

For example if I have Integral from 0 to 1 of x (2 + 3x^2) dx
But it doesn't change the explication ^^