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a substitution will do, let =3x

let u=3x*

When I solved it, I got:

can you go from here?

I don't know what to do after this..

you integrate and backsubstitute. you will get:
\[ \sqrt[3]{u}^2\]

So that's the answer?

the answer is
\[ \sqrt[3]{3x}^2 \]

What do you do after you integrate and back substitute? I want to show all the steps.

Then you are done. That's the indefinite integral.

But the answer is cube root of 3x^2. You did something after you got cube root of u^2.

Oh okay, can you help me with one more?

sure

Do you know basic integration of exponents yes?

Some of it.

because you can write this integral like this:
\[ \int (x^{-2} -x^{-3})dx\]

And then you just integrate normally, increase the exponent by one and divide by the new exponent.

Can we do it step by step?

yes one second

Okay

if you integrate the first term you should get \[- \frac{1}{x}\]

Okay, I got that part.

think you can figure out the second one? it works exactly the same.

Wait, I mean -1/2x^2

yes you did it right

It's the first or second one?

\[ \frac{x^{-3+1}}{-3+1} = -\frac{1}{2x^2}\]

Okay, what is the step after that?

So it would be: -1/2x^2+C?

-1/x - 1/2x^2 + C = Answer?

\[ \int x^{-2} dx- \int x^{-3}dx = -\frac{1}{x} + \frac{1}{x^2}+C \]

almost, the minus becomes a plus, because there is a minus in front of the integral. See it?

oops, you are absolutely right, I messed it up

\[ \int x^{-2} dx- \int x^{-3}dx = -\frac{1}{x} + \frac{1}{2x^2}+C \]

well done! (-: you discovered my mistake.

Yay! I need help with a few more if you have time?

I believe so. just post them.

you know how to setup this integral?

No

I'm not good with trigs..at all :/

hmm well I more meant, do you know what you have to do in order to calculate the area of this ?

No

The integral calculates the Area below a graph, do you know that?

Okay

So do you have the picture open?

Yes, I have it open

0 to pi/2

exactly!

and the function is y=x+cosx right?

Yes

Okay, that's all there is to it?

Sorry, I'm so confused. This is due by 9:30 and I'm under stress. Ugh!

ohhh, I see.

So, now I plug in pi/2 and then plug in 0?

first you need to compute the antiderivate.

have you done that? can you tell me what the integral of x is ?

x^2/2

exactly!

That's the difficult part. Wouldn't I need a unit circle?

Wouldn't it be negative sin(x)?

no the derivative of cos(x) is -sin(x)

but we are doing integration , that's the other way around.

Okay, now I just plug in pi/2 into sin x?

I get sin(pi/2) = 1

that's true!

so the entire area is?

-1 ?

So it's 1.

yes, but that's not the entire area

Remember what we got for the first integral?

Now I have to subtract pi/8+1 by 1?

no, that's the solution already up there, I did underline it.

Ohhh so the entire area is pi/8 +1.

Yes so how would I write the final answer?

just like I did, up there

\[ A
= \underline{\underline{\frac{\pi^2}{8}+1}}\]

because that is the area.

we can give it a try.

one you answered right

How do I explain it in sentences though?

two you answered right two, but your reasoning isn't quite correct

If I use partial integration, how would it look?

\[ \large\int f'(x)g(x)dx= f(x)g(x)- \int f(x)g'(x) dx \]

Okay

For 3-5, I just don't know how to word it.

false because y= ln e =1 is a number, the derivative of a number is always zero.

That's all I have to say for that?

totally

for 3.

Okay, great, 4 and 5 left.

4 and 5 you answered perfectly already.

there is nothing I would add.

But I have to rephrase it all into sentence form and I don't know how.

ah yeh? Well, this is math, sometimes showing the example is the best way to make a phrase.

Okay, thank you so so much!!

welcome