I'm doing some integral exercises at the moment. Anyone want to verify that my solution is correct as seen in the following link?

https://pasteboard.co/Hgx8tXQ.png

(April 14, 2018 at 2:07 am)Grandizer Wrote: I'm doing some integral exercises at the moment. Anyone want to verify that my solution is correct as seen in the following link?

https://pasteboard.co/Hgx8tXQ.png

It's good.

(April 14, 2018 at 2:07 am)Grandizer Wrote: I'm doing some integral exercises at the moment. Anyone want to verify that my solution is correct as seen in the following link?

https://pasteboard.co/Hgx8tXQ.png

Yup looks good!

I'd advise keeping the brackets in place after putting in the limits, as an extra step, to reduce the risk of dropping a minus sign somewhere:

(3/5 + 6 + 16) - (-3/5 + 6 - 16)

But you didn't drop any

Yeah, the beauty of "writing" this out on a computer program instead of on paper is that I can easily remove some steps once I'm sure I got them right (just to have some space and/or to make the presentation look neater).

Anyway, just now I worked on the following problem:

Find the area of the region between the curves of the functions x^-2 and -x^-3, over the interval x>1 ... which is the same as the interval (1,infinity).

Here's my solution:

https://pasteboard.co/HgHPq36.png

Hopefully I got it right.

(April 15, 2018 at 5:18 am)Grandizer Wrote: ...
Hopefully I got it right.

yes, clear correct solution

Yup, looks all good to me!

This kind of question shows the awesome power of mathematics (and specifically calculus). Imagine trying to approach this without it.

For the indefinite integral evaluation, forgot to include the constant c. Not a big deal as the final answer is unaffected by not including it, but still a mistake.

(April 15, 2018 at 9:21 am)Grandizer Wrote: For the indefinite integral evaluation, forgot to include the constant c. Not a big deal as the final answer is unaffected by not including it, but still a mistake.

I would put the integral calculation inline with the limit. In that way, no constant is required. Just the evaluation from 1 to q.

The indefinite integral of 1/x is ln|x|+c ... x!=0.

Here's to show how this works for negative values of x, keeping in mind that the indefinite integral is the antiderivative in general.

When x is negative, |x| = -x.

d/dx(ln|x|+c) = (1/(-x))*(-1)+0 = 1/x

(March 8, 2018 at 3:49 am)robvalue Wrote: and it shows what I was saying about how understanding helps the memory. It all follows, it's not just meaningless drivel to memorise.

That's why I struggle with remembering phone numbers.