How can I find the indefinite integral of the attached problem?

- anonymous

How can I find the indefinite integral of the attached problem?

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- anonymous

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- anonymous

a substitution will do, let =3x

- anonymous

let u=3x*

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- anonymous

When I solved it, I got:

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- anonymous

\[ u = 3x \longrightarrow \frac{du}{dx}=3
\\
\frac{2}{3}\cdot \frac{du}{dx}=2 \]
\[ \large \frac{2}{3}\int u^{-\frac{1}{3}}du\]

- anonymous

can you go from here?

- anonymous

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

- anonymous

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

- anonymous

So that's the answer?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

well at the beginning I said substitute u=3x, and you want your result to be in forms of x right? so you have to back-substitute, where there is an u, you put a 3x instead.

- anonymous

Oh okay, can you help me with one more?

- anonymous

sure

- anonymous

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- anonymous

Do you know basic integration of exponents yes?

- anonymous

Some of it.

- anonymous

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

- anonymous

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

- anonymous

Can we do it step by step?

- anonymous

yes one second

- anonymous

Okay

- anonymous

so you can actually split this integral up if that makes it easier for you
\[ \int x^{-2} dx- \int x^{-3}dx \]

- anonymous

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

- anonymous

Okay, I got that part.

- anonymous

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

- anonymous

Wait, I mean -1/2x^2

- anonymous

yes you did it right

- anonymous

It's the first or second one?

- anonymous

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

- anonymous

Okay, what is the step after that?

- anonymous

That's it, you have taken the indefinite integral so you can just write it down. usually you add a + C at the end for indefinite integrals to show that an arbitrary constant could have been lost.

- anonymous

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

- anonymous

no, both of them, you have taken two indefinite integrals remember? Look at the formula I have posted above, we did split them up, you have derived the first one and the second one, both are part of your solution.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Yes, but wasn't the second part -1/ 2x^2? Then in the answer, why did you leave out the 2 in front of the x^2?

- anonymous

oops, you are absolutely right, I messed it up

- anonymous

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

- anonymous

well done! (-: you discovered my mistake.

- anonymous

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

- anonymous

I believe so. just post them.

- anonymous

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- anonymous

you know how to setup this integral?

- anonymous

No

- anonymous

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

- anonymous

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

- anonymous

No

- anonymous

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

- anonymous

So whenever you want to calculate the area underneath a graph of a function, of any kind, you want to setup an integral.

- anonymous

\[ \large \int_a^b f(x) dx \] this is the general setup of an integral, it is how it looks like, y=f(x) is the function you want to integrate, and the dx just tells you in respect to what you have to integrate it with.

- anonymous

a and b are the boundaries of your integral, more correctly it should be written like this
\[ \int_{x=a}^{x=b} f(x)dx\]

- anonymous

for a < b, that means you integrate a function, starting from the point a, and you integrate it's entire area until you hit the point b.

- anonymous

the under bound is always the lower limits, meaning lower value, the upper bound is always the upper limit, meaning higher value.

- anonymous

Okay

- anonymous

So do you have the picture open?

- anonymous

Yes, I have it open

- anonymous

So can you tell me what the boundaries (upper and lower) for this integral are?, from where to where does it run? Look at the area you want to integrate.

- anonymous

0 to pi/2

- anonymous

exactly!

- anonymous

and the function is y=x+cosx right?

- anonymous

Yes

- anonymous

so this means it looks like that :
\[ \large \int_0^{\frac{\pi}{2}}(x + \cos x)dx =\int_0^{\frac{\pi}{2}}xdx + \int_0^{\frac{\pi}{2}}\cos(x)dx \]

- anonymous

Both are easy integrals for themselves, when you look at them individually, keep in mind, the integrals turns a function into it's antiderivative, that means if you derive an integrated function, you should get back to the original function.

- anonymous

Okay, that's all there is to it?

- anonymous

well now you have to evaluate it, you want to determine the area of the given region, that means you want to compute the integral and evaluate it.

- anonymous

just like you did before, take the antiderivative, only this time you have values to plug in and see what happens.

- anonymous

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

- anonymous

ohhh, I see.

- anonymous

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

- anonymous

first you need to compute the antiderivate.

- anonymous

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

- anonymous

x^2/2

- anonymous

exactly!

- anonymous

and see, there you plug in the values now, it always works like this, plug in the higher value, then subtract from that the value that you get when you plug in the lower value.

- anonymous

Usually you see this written like that
\[ \left. \frac{x^2}{2} \right|_{x=0}^{x=\frac{\pi}{2}}\ = \frac{\frac{\pi ^2}{4}}{2} - \frac{0^2}{2} = \frac{\pi^2}{8} \]

- anonymous

but that's only the first step, now you need to do the exact same with the cos(x) function, your second integral.

- anonymous

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

- anonymous

no, you need to know the trig functions and their derivative by heart
the derivative of sin(x) = cos(x)
therefore the antiderivative of cos(x) = sin(x)

- anonymous

Wouldn't it be negative sin(x)?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

exactly and then you subtract from that sin(0) , but sin(0) =0, so you don't have to worry about that, just to empathize the order.

- anonymous

I get sin(pi/2) = 1

- anonymous

that's true!

- anonymous

so the entire area is?

- anonymous

-1 ?

- anonymous

no you add them both together now, just like we did before with the indefinite integrals, now you add the values together

- anonymous

So it's 1.

- anonymous

yes, but that's not the entire area

- anonymous

Remember what we got for the first integral?

- anonymous

\[\large \int_0^{\frac{\pi}{2}}(x + \cos x)dx =\int_0^{\frac{\pi}{2}}xdx + \int_0^{\frac{\pi}{2}}\cos(x)dx \\
= \underline{\underline{\frac{\pi^2}{8}+1}}\]

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

see \(\frac{\pi^2}{8}\) is what we got for the first integral, when we evaluated it, remember? x^2/2, you did that yourself, now the next integral - it has a plus there, you evaluated that integral and you said the area is 1
so the entire area is both summed together.

- anonymous

Yes so how would I write the final answer?

- anonymous

just like I did, up there

- anonymous

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

- anonymous

because that is the area.

- anonymous

Okay, thank you so much! Are you good with true and false questions? I have the answers already but I have to write sentences and I need help with that.

- anonymous

we can give it a try.

- anonymous

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- anonymous

one you answered right

- anonymous

How do I explain it in sentences though?

- anonymous

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

- anonymous

False: The integral of two functions cannot be computed by taking the integral of each function individually and multiplying the results. The correct method is using partial integration.

- anonymous

If I use partial integration, how would it look?

- anonymous

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

- anonymous

Okay

- anonymous

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

- anonymous

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

- anonymous

That's all I have to say for that?

- anonymous

totally

- anonymous

for 3.

- anonymous

Okay, great, 4 and 5 left.

- anonymous

4 and 5 you answered perfectly already.

- anonymous

there is nothing I would add.

- anonymous

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

- anonymous

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

- anonymous

You can always make sentences though of course, for 4 you can write. False, if you derive the natural log of x you get the inverse function, when you integrate it you again have to use the technic of partial integration which leads to xlnx-x + C

- anonymous

For 5. There is a function which is equal to it's derivative, this function is e^x, it's slope at any given point of x is equal to it's function valued at given point.

- anonymous

Okay, thank you so so much!!

- anonymous

welcome

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