Question

find the indefinite integral x^-13 dx

Answer 1

when you are integrating a power series you add 1 to the exponent and divide by that exponent, and make to to add C since its indefinite

Answer 2

\[\bf \int\limits_{}^{}x^ndx=\frac{ x^{n+1} }{ n+1 }\]\[\bf \int\limits_{}^{}x^{-13}dx=? \]

Answer 3

+ C; I forgot the constant.

Answer 4

x^-12/-12+C

Answer 5

Correct.

Answer 6

Awesome!! Thank you!!

Answer 7

$$\begin{align*}\int x^{-13}\,\mathrm{d}x&=-\frac1{12}\int\left(-12x^{-13}\right)\,\mathrm{d}x\\&=-\frac1{12}\int\frac{\mathrm{d}}{\mathrm{d}x}\left(x^{-12}\right)\,\mathrm{d}x\\&=-\frac1{12}x^{-12}+C\end{align*}$$

Answer 8

@oldrin.bataku Don't give away answers.

Answer 9

It's not quite giving away an answer if it was already previously posted, not to mention I accompanied it with work that demonstrate why the rule works.$$\int x^n\,dx=\frac1{n+1}\int((n+1)x^n)\,dx=\frac1{n+1}\int\frac{d}{dx}\left(x^{n+1}\right)\,dx=\frac1{n+1}x^{n+1}+C$$

Answer 10

The rule works because you are inversely differentiating..If:\[\bf f(x)=x^n \ then \ f'(x)=nx^{n-1}\], then inversely differentiating, i.e. adding 1 to the exponent and then dividing by this new exponent brings you to the antiderivative.

Answer 11

This is obviously not an argument, just a complement to what you've already posted.