Question

Evaluate the integral, small question on formula/theory/concept.

Answer 1

How does 2dx become 2x --to be precise?

Answer 2

joke?

\[\int_{-11}^{11}2\,dx=2\int_{-11}^{11}dx=\left.2x\right|_{-11}^{11}.\]

Answer 3

I'm just wondering what happened to the 'd'

Answer 4

o_o

The integral you posted is seriously complicated. I'd guess you know what \(dx\) stands for, since that is trivial as heck.

Answer 5

More precisely, \(dx\) is a basis element of the cotangent bundle of \(\mathbb R\) with the standard topology.

Answer 6

You could look at it the other way, start from this derivative
\[\frac{d}{dx}( 2x) = 2\]
Integrate both sides,
\[\int_{-11}^{11} \frac{d}{dx} ( 2x) dx = \int_{-11}^{11} 2 dx\]
The left hand side simplifies because of the funamental theorem of calculus,
\[2x |_{-11}^{11} = \int_{-11}^{11} 2 dx\]

Answer 7

Ah, basically the anti-derivative. I got it now, just a small question :)

Answer 8

Thanks @Kainui and @across

Answer 9

I need to learn how to teach, xD.

Answer 10

Yeah I think trouble with integrating 2 and saying it's trivial and mentioning tangent bundles is probably a bad idea lol xD

Answer 11

\[\large\boxed{\boxed{\int_a^bax^n\,\mathrm dx=\left.\frac{ax^{n+1}}{n+1}\right|_a^b}}\]
\[\int_{-11}^{11}2\,\mathrm dx
=\int_{-11}^{11}2x^0\,\mathrm dx
=\left.\frac{2x^{0+1}}{0+1}\right|_{-11}^{11}
=\left.2x\right|_{-11}^{11}\]