# Evaluating definite integrals via substitution.

#### shamieh

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

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

#### MarkFL

Staff member
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
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

##### Well-known member
As you have not changed $$\displaystyle x$$ to $$\displaystyle u$$ you can't change limts from $$\displaystyle \int_0^1$$ to $$\displaystyle \int_1^2$$

#### shamieh

##### Active member
As you have not changed $$\displaystyle x$$ to $$\displaystyle u$$ you can't change limts from $$\displaystyle \int_0^1$$ to $$\displaystyle \int_1^2$$
?

- - - Updated - - -

Anyone else getting $$\displaystyle \frac{708204}{5}$$ ?

#### ZaidAlyafey

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

#### shamieh

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

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
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
Awesome, thanks.

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