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Evaluating definite integrals via substitution.

shamieh

Active member
Sep 13, 2013
539
Can someone make sure I'm on the right track with this problem? I'm a little confused because I thought that when you make a substitution you update the limits and get better numbers to work with when you plug them in the function in the end...Yet, it seems like I almost got worse numbers to work with.. Here is the problem, and what I have done so far.

Evaluate the following definite integral.

\(\displaystyle \int^1_0 36x^2(x^3 + 1)^4\)

\(\displaystyle u = x^3 + 1\)
\(\displaystyle du = 3x^2\)

But I don't have a \(\displaystyle 3x^2\) up top, I have a \(\displaystyle 36x^2\) so I divided out and got

\(\displaystyle \frac{du}{3} = x^2\)

now I update the limits and I get

\(\displaystyle 0^3 + 1 = 1\)
\(\displaystyle 1^3 + 1 = 2\)
so

\(\displaystyle \frac{1}{3} \int^2_1 du * u^4 = \frac{1}{3} * \frac{1}{5}u^5 = \frac{1}{15} * (2^3 + 1)^4] - [\frac{1}{15} * (1^3 + 1)^4]\)

Does this look correct? or have i messed up somewhere? I mean really i have to do \(\displaystyle 9^4\)? (sorry if that sounds ignorant, just want to make sure)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \frac{1}{3} \int^2_1 du * u^4 = \frac{1}{3} * \frac{1}{5}u^5 = \frac{1}{15} * (2^3 + 1)^4] - [\frac{1}{15} * (1^3 + 1)^4]\)

\(\displaystyle 36 \, \int^2_1 \frac{du}{3}\cdot \, u^4 =12 \int^2_1 \, du \cdot u^4\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would write the integral as:

\(\displaystyle 12\int_0^1\left(x^3+1 \right)^4\,3x^2\,dx\)

to get the result obtained by ZaidAlyafey.
 

shamieh

Active member
Sep 13, 2013
539
Seems I forgot to include the original 36 in the problem.

so I ended up with this:

\(\displaystyle [\frac{12}{5} (2^3 + 1)^5] - [\frac{12}{5} (1^3 + 1)^5]\)

\(\displaystyle [\frac{12}{5} * 59,049] - [\frac{12}{5} * 32]\)

\(\displaystyle [\frac{708,588}{5}] - [\frac{384}{5}] = \frac{708,204}{5}\)

This doesn't seem right though...Can anyone check my work?
 

mathworker

Active member
May 31, 2013
118
As you have not changed \(\displaystyle x\) to \(\displaystyle u\) you can't change limts from \(\displaystyle \int_0^1\) to \(\displaystyle \int_1^2\):rolleyes:
 

shamieh

Active member
Sep 13, 2013
539
As you have not changed \(\displaystyle x\) to \(\displaystyle u\) you can't change limts from \(\displaystyle \int_0^1\) to \(\displaystyle \int_1^2\):rolleyes:
?:confused:

- - - Updated - - -

Anyone else getting \(\displaystyle \frac{708204}{5}\) ?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle 12 \int^2_1 \, du \cdot u^4 = \frac{12}{5} \left( 2^5-1\right) = \frac{31 \cdot 12}{5}=\frac{372}{5}\)
 

shamieh

Active member
Sep 13, 2013
539
\(\displaystyle 12 \int^2_1 \, du \cdot u^4 = \frac{12}{5} \left( 2^5-1\right) = \frac{31 \cdot 12}{5}=\frac{372}{5}\)

Zaid, is it because I changed the limits that I don't have to say, for example, [higher limit in the function] - [the lower limit in the function?]

In any other limit problem I would say [maximum limit plugged into antiderivative ] - [minimum plugged into antiderivative].. now you are just saying [maximum number plugged into anti derivative?]
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Well, if you want to be in the safe side just find the anti-derivative. which is in your case

\(\displaystyle 12\frac{(x^3+1)^5}{5}+C\)

It remains just to use the FTC.
 

shamieh

Active member
Sep 13, 2013
539
Awesome, thanks.

Instead of

\(\displaystyle \frac{12}{5}(2)^5 - \frac{12}{5} (1)^5\)

I was doing \(\displaystyle \frac{12}{5}(2^3 + 1)^4\) - ....(Dull)