Integration with term outside brackets

\displaystyle \int^2_0 z(z+5)\ dz

I thought that I would isolate the z coeffecient and brackets, to give me, z^2, 5z^2/2

However, according to an online integral calcuator, the correct answer is 2z^3+15z^2/6

Im sorry; but, the latex wont let me put down 20 and 10 for the integration, that is what the original question is asking.

Where am I going wrong?

(edited 9 years ago)

Reply 1

TenOfThem

Original post by apronedsamurai

\displaystyle \int^{20}_{10} z(z+5)\ dz

I thought that I would isolate the z coeffecient and brackets, to give me, z^2, 5z^2/2

However, according to an online integral calcuator, the correct answer is 2x^3+15x^2/6

Im sorry; but, the latex wont let me put down 20 and 10 for the integration, that is what the original question is asking.

Where am I going wrong?

Are you integrating wrt to z

Why is the x there

(edited 9 years ago)

Reply 2

User947387

Sorry that was a typo. What is wrt?

Reply 3

TenOfThem

Original post by apronedsamurai

Sorry that was a typo. What is wrt?

If we get rid of the x are we integrating dz

Look at the corrected version I have posted in post 2 is that what you are integrating

If so multiply the bracket out and integrate as normal

(edited 9 years ago)

Reply 4

User947387

Sorry, again, the all the x's were mistakes. Z's

Reply 5

TenOfThem

Original post by apronedsamurai

Sorry, again, the all the x's were mistakes. Z's

Again

Please look at what I have put in post 2 is that what you want to integrate

Yes, it is.

Original post by apronedsamurai

Yes, it is.

Then

As I said

Multiply out the bracket and integrate as normal

You can not just integrate the separate bits

Reply 8

User947387

Multiplying out the brackets (before integration) yields me z^2+5z?

(edited 9 years ago)

Reply 9

User947387

Would I need to use the substitution method here?

Reply 10

TenOfThem

Original post by apronedsamurai

Multiplying out the brackets (before integration) yields me z^2+5z?

Original post by apronedsamurai

Would I need to use the substitution method here?

Why do you think you need to substitute

Increase the power and divide

Btw if you use the reply button on my post I will get a notification that you have replied to me

Reply 11

User947387

That is what I am confused by. Do i raise the power of the Z coeffiecent (the Z term outside the brackets) or do I multiply out the brackets, THEN raise the power by one?

Reply 12

TenOfThem

Original post by apronedsamurai

That is what I am confused by. Do i raise the power of the Z coeffiecent (the Z term outside the brackets) or do I multiply out the brackets, THEN raise the power by one?

I have answered that question twice already

You cannot apply integration rules to the separate parts

You have multiplied the bracket - NOW you integrate

Reply 13

User947387

Thank you very much.

I got the correct answer (3083 and a third) :smile:

I was thrown by the online calculator, as it had gave such a wildly different result, with regards to the integration of the terms :| Perhaps that was just an input error, or I interperted the information incorrectly.

Regardless, I have now taken another step towards understanding integration. Thank you.