# substitution method question

• Apr 9th 2006, 10:22 PM
cinder
substitution method question
Do can you tell when you can use the substitution method? I just tried it on $\displaystyle \int_{-1}^0(2x+3)^2dx$, but I don't get the right answer unless I square the $\displaystyle (2x+3)$ and then proceed normally.

What about substitution on $\displaystyle \int_3^2\frac{x^2-1}{x-1}dx$?
• Apr 10th 2006, 01:05 AM
earboth
Quote:

Originally Posted by cinder
Do can you tell when you can use the substitution method? I just tried it on $\displaystyle \int_{-1}^0(2x+3)^2dx$, but I don't get the right answer unless I square the $\displaystyle (2x+3)$ and then proceed normally.
What about substitution on $\displaystyle \int_3^2\frac{x^2-1}{x-1}dx$?

Hello,

to 1.:
use u(x) = 2x+3. Then du/dx = 2. That means du = 2*dx.

So you could use the substitution method if there are the factor 2:

$\displaystyle \int_{-1}^0 \frac{1}{2} \cdot 2 \cdot (2x+3)^2dx=\int_{-1}^0 \frac{1}{2} u^2 du$

$\displaystyle \int_{-1}^0 \frac{1}{2} u^2 du= \frac{1}{6}u^3+c$

Now you can re-substitute and you'll get:

$\displaystyle \int_{-1}^0 \frac{1}{2} \cdot 2 \cdot (2x+3)^2dx=\frac{1}{6} \cdot (2x+3)^3+c$

to 2.:
Factorize (x^2-1) = (x+1)(x-1). Then you can cancel (x-1) and you'll get:

$\displaystyle \int_3^2\frac{x^2-1}{x-1}dx= \int_3^2 (x+1)dx$.

From here on you know what to do!

Greetings

EB
• Apr 10th 2006, 06:10 AM
cinder
Quote:

Originally Posted by earboth
Hello,

to 1.:
use u(x) = 2x+3. Then du/dx = 2. That means du = 2*dx.

So you could use the substitution method if there are the factor 2:

$\displaystyle \int_{-1}^0 \frac{1}{2} \cdot 2 \cdot (2x+3)^2dx=\int_{-1}^0 \frac{1}{2} u^2 du$

$\displaystyle \int_{-1}^0 \frac{1}{2} u^2 du= \frac{1}{6}u^3+c$

Now you can re-substitute and you'll get:

$\displaystyle \int_{-1}^0 \frac{1}{2} \cdot 2 \cdot (2x+3)^2dx=\frac{1}{6} \cdot (2x+3)^3+c$

to 2.:
Factorize (x^2-1) = (x+1)(x-1). Then you can cancel (x-1) and you'll get:

$\displaystyle \int_3^2\frac{x^2-1}{x-1}dx= \int_3^2 (x+1)dx$.

From here on you know what to do!

Greetings

EB

Okay, so on the first one, could you do:

$\displaystyle \int_{-1}^0(2x+3)^2dx$

$\displaystyle u = (2x + 3)$
$\displaystyle du = 2dx$ which becomes $\displaystyle \frac{1}{2}du=dx$

then

$\displaystyle \int_1^3u^2\frac{1}{2}du = \frac{1}{2}\int_1^3u^2du$

and solve normally, just using $\displaystyle du$?
• Apr 10th 2006, 06:52 AM
ThePerfectHacker
Quote:

Originally Posted by cinder
and solve normally, just using $\displaystyle du$?

Yes, that what $\displaystyle du$ means, a function in terms of $\displaystyle u$ whose derivative gives you whatever the function is. Or another way of saying the antiderivative in terms of $\displaystyle u$. That is the entire point of substitution because it simplifys the integrand into a simpler one.
• Apr 10th 2006, 07:27 AM
cinder
Quote:

Originally Posted by ThePerfectHacker
Yes, that what $\displaystyle du$ means, a function in terms of $\displaystyle u$ whose derivative gives you whatever the function is. Or another way of saying the antiderivative in terms of $\displaystyle u$. That is the entire point of substitution because it simplifys the integrand into a simpler one.

Okay, did I do the substitution right? :confused:
• Apr 10th 2006, 07:30 AM
ThePerfectHacker
Quote:

Originally Posted by cinder
Okay, did I do the substitution right? :confused:

Yes
• Apr 10th 2006, 07:33 AM
cinder
Quote:

Originally Posted by ThePerfectHacker
Yes

Thanks. :)