Question

@zepdrix

Answer 1

Integral of \(\Large 4^{-x}\)

Answer 2

Ooo there's a tricky one huh :D

Answer 3

\[\large\rm \int\limits 4^{-x}dx\]

Do you remember this derivative?\[\large\rm \frac{d}{dx}a^x=?\]

Answer 4

yes, the derivative of a^x = a^x(ln x)(dx)

Answer 5

by dx I mean the derivative of x

Answer 6

woops!
we multiply by the log of the BASE :)\[\large\rm \frac{d}{dx}a^x=a^x(\ln a)\]

Answer 7

Oh, right

Answer 8

So if we integrate \(\large\rm a^x(ln a)dx\)
we have to `divide out` the lna to end up with the correct answer, right?\[\large\rm \int\limits a^x(\ln a)dx=a^x\]So then,\[\large\rm \int\limits a^x dx=\frac{a^x}{\ln a}\]

Answer 9

And again, maybe for this problem, do a u-substitution to deal with the - on the x.
u=-x

Answer 10

yes, but there is no ln a in the problem...

Answer 11

Think about what integration is doing.
Taking derivative told you to simply `multiply by (ln a)`.
So then taking integral means you simply `divide by (ln a)`.

When you take derivative, you `gained` an (ln a).
So when you integrate, you `lose` an (ln a).

Answer 12

We can be more fancy about it if you prefer,
hmm thinking :)

Answer 13

Oh, ok.

Answer 14

I understand that now.

Answer 15

Here is another idea,
we multiply our integral by (ln a)/(ln a).\[\large\rm \int\limits \frac{a^x(\ln a)}{\ln a}dx\]We'll pull the bottom one out front, because it's an unloved stupid coefficient :)\[\large\rm \frac{1}{\ln a}\cdot\int\limits a^x(\ln a)dx\qquad=\qquad \frac{1}{\ln a}\cdot a^x\]

Answer 16

Okay, and do we also then have to divide by the anti derivative of the exponent?

Answer 17

Yesssss, good good good.
This trick of `dividing out the coefficient` only works when our x is `linear`.
If we had like x^2 in the exponent,
it would be a whole nother story!

But yes, when it's just a linear x,
we can do these little division tricks.

Answer 18

so the antiderivative of -x is -x^2/2, right?

Answer 19

This will work for all kinds of linear coefficients,
here is another example:\[\large\rm \int\limits \cos(2x)dx\quad=\quad \frac12\sin(2x)\]
Derivative would normally `give us` an extra 2,
so through integration `we lose` a 2.

Answer 20

Ooo my bad, I explained that bad I guess :)
No, we're not taking the anti-derivative of -x.
We're simply dividing by the coefficient -1 to compensate for chain rule.

Answer 21

Oh, so then the answer would be \(\Large -\frac{4^{-x}}{\ln x}\) + C?

Answer 22

Good good good :)
But again, try to put that exponent derivative to memory,
`log of base`, not log of x.

Answer 23

Ugh, so \(\Large -\frac{4^{-x}}{\ln 4}\)+ C?

Answer 24

yay good job \c:/

Answer 25

Thank you!

Answer 26

np