Studying Mathematics Thread

[GrandizerII]

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

[polymath257]

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.

[robvalue]

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

[GrandizerII]

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.

[Jarrey]

...
Hopefully I got it right.

yes, clear correct solution

[robvalue]

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.

[GrandizerII]

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.

[polymath257]

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.

[GrandizerII]

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

[Edwardo Piet]

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.