Question

Integrate :-

Answer 1

@genius12 @Jack199

Answer 2

$$\int a^{-x} dx=-\frac{a^{-x}}{\ln(a)}+C, \text{ for } a \ne 1$$
$$\int a^{-x}\ dx=x+C, \text{ for } a=1$$

Answer 3

Let \(\bf u = -x \implies dx=-du\). Then:\[\bf =-\int\limits_{}^{}a^u \ du\]And use:\[\bf \int\limits_{}^{}a^x dx=\frac{ a^x }{ \ln(a) }+C, \ a \ne 1\]

Answer 4

Why are you using boldface for your variables, they look like vectors

Answer 5

And: \[\bf \int\limits_{}^{}a^x \ dx=x+C, \ a=1\]

Answer 6

Get it into an integrable form

Let y = a^-x
\[\Large y = a^{-x}\]
take logs of both sides
\[\Large \ln y = \ln a^{-x}\]
simplify with log laws
\[\Large \ln y = -x \ln a\]
put both sides to the power of e
\[\Large y = e^{-x lna}\]
Now put that back into your integral

\[\huge \int\limits\limits e^ {-xln a} dx\]

Answer 7

@agent0smith A u-substitution is much easier..

Answer 8

I would be inclined to say agent0smith's technique is easyier, he just wrote it out all the way