# Thread: integrals - substitution rule: some clarification

1. ## integrals - substitution rule: some clarification

Heres the problem and the issue i need help with circled in red:

Im trying to figure out how exactly you are supposed to calculate du. I see that i take the derivative of u, but im unsure how to go about it from there on. Im having the same issue in other problems and some clarification would be much help. Thanks.

2. Originally Posted by Evan.Kimia
Heres the problem and the issue i need help with circled in red:

Im trying to figure out how exactly you are supposed to calculate du. I see that i take the derivative of u, but im unsure how to go about it from there on. Im having the same issue in other problems and some clarification would be much help. Thanks.
$\displaystyle u = -0.01t \Rightarrow \frac{du}{dt} = -0.01 \Rightarrow du = -0.01 dt \Rightarrow dt = - \frac{1}{0.01} du = -100 du$.

3. Originally Posted by Evan.Kimia
Heres the problem and the issue i need help with circled in red:

Im trying to figure out how exactly you are supposed to calculate du. I see that i take the derivative of u, but im unsure how to go about it from there on. Im having the same issue in other problems and some clarification would be much help. Thanks.
Dear Evan.Kimia,

$\displaystyle u=-0.01t$

From the definition of differentials, $\displaystyle du=-0.01dt$

Refer:Elementary Calculus: Definition of Differentials

4. thank you both, i realized i just didnt notice that they simplified it.
I did have another question related to solving for dx for another problem:

when solving for dx this is my method:
$\displaystyle u=-x^{30}$

$\displaystyle du=-30x^{29}$

$\displaystyle \frac{du}{dx}=-30x^{29}$

$\displaystyle du=-30x^{29}dx$

$\displaystyle dx=-\frac{1}{30x^{29}}du$

would this calculation be correct? Because if it is, im not sure how to continue from here.

5. Originally Posted by Evan.Kimia
thank you both, i realized i just didnt notice that they simplified it.
I did have another question related to solving for dx for another problem:

when solving for dx this is my method:
$\displaystyle u=-x^{30}$

$\displaystyle du=-30x^{29}dx$

$\displaystyle \frac{du}{dx}=-30x^{29}$

$\displaystyle du=-30x^{29}dx$

$\displaystyle dx=-\frac{1}{30x^{29}}du$

would this calculation be correct? Because if it is, im not sure how to continue from here.
Yes.

Then the integration becomes:

$\displaystyle \int_0^1 x^{29} e^{u} \frac{-du}{30x^{29}} = \frac{-1}{30}\int_0^1 e^u du$

now integrate, and then substitute the value of u, then solve for the given upper and lower limits

6. Originally Posted by Evan.Kimia
thank you both, i realized i just didnt notice that they simplified it.
I did have another question related to solving for dx for another problem:

when solving for dx this is my method:
$\displaystyle u=-x^{30}$

$\displaystyle du=-30x^{29}$

$\displaystyle \frac{du}{dx}=-30x^{29}$

$\displaystyle du=-30x^{29}dx$

$\displaystyle dx=-\frac{1}{30x^{29}}du$

would this calculation be correct? Because if it is, im not sure how to continue from here.
Dear Evan.Kimia,

Your calculation is correct.(although I assume that the second line of your calculation is a typo.) However you could continue from here by substituting what you obtained for dx in the original equation.

7. Originally Posted by harish21
Yes.

Then the integration becomes:

$\displaystyle \int_0^1 x^{29} e^{u} \frac{-du}{30x^{29}} = \frac{-1}{30}\int_0^1 e^u du$

now integrate, and then substitute the value of u, then solve for the given upper and lower limits
It is better to confirm that the OP has the correct expression for dx and then let him/her continue with the question rather than posting virtually the rest of the solution. However, there is more left to do than meets the eye - because there is an error in your work which I would prefer ws not pointed out (leave something more for the OP to do).

8. Which part would be considered a typo? thanks.

9. Originally Posted by Evan.Kimia
Which part would be considered a typo? thanks.
While solving for dx, you have written this :

$\displaystyle u=-x^{30}$

$\displaystyle du=-30x^{29}$

but this is supposed to be stated as

$\displaystyle \frac{du}{dx} = -30x^{29}$

or, $\displaystyle du = -30x^{29} dx$

10. Originally Posted by harish21
While solving for dx, you have written this :

$\displaystyle u=-x^{30}$

$\displaystyle du=-30x^{29}$

but this is supposed to be stated as

$\displaystyle \frac{du}{dx} = -30x^{29}$

or, $\displaystyle du = -30x^{29} dx$
That is not the error. As to where the error is, the OP is encouraged to work through the question him/herself, using what has been learned in this thread ....

11. Originally Posted by Evan.Kimia
Which part would be considered a typo? thanks.
Dear Evan.Kimia,

I meant the second line of your post...

$\displaystyle {\color{red}du=-30x^{29}}$ it should be, $\displaystyle du=-30x^{29}dx$

12. Originally Posted by Sudharaka
Dear Evan.Kimia,

I meant the second line of your post...

$\displaystyle {\color{red}du=-30x^{29}}$ it should be, $\displaystyle du=-30x^{29}dx$
That is not the error I was referring to. I was referring to an error in this:

Originally Posted by Sudharaka

$\displaystyle \int_0^1 x^{29} e^{u} \frac{-du}{30x^{29}} = \frac{-1}{30}\int_0^1 e^u du$