#1

Hate to start such a lame thread but does anyone know how to integrate y = a^x. I used to know the answer but I'm not sure I remember it right and I can't find it anywhere.

Cheers.

Cheers.

#2

it is 3

#3

dy/dx = xa^x-1?

Something like that. Or is that differentiation?

Something like that. Or is that differentiation?

#4

Integrated it is (a^x+1)/x+1

#5

(1/x+1)a^(x+1)

I think.

If you differentiate it you get

1a^x

So I'm pretty sure it's right.

EDIT: Or you can put it like BadReligionRock did.

I think.

If you differentiate it you get

1a^x

So I'm pretty sure it's right.

EDIT: Or you can put it like BadReligionRock did.

#6

I'm not sure about that. I seem to remember that there was a standard formula where the answer involved logarithms.

#7

I'm not sure about that. I seem to remember that there was a standard formula where the answer involved logarithms.

Where you put "a", you don't mean what we see as "e" - as in the exponential function, do you?

#8

I'm not sure about that. I seem to remember that there was a standard formula where the answer involved logarithms.

What I've shown you is right. Add one to the power, and divide by the new power.

#9

Isn't that what you'd do if it was x^a? You can't do that with a^x can you?

#10

a^x = e^ln(a^x) (as the e and ln will cancel each other.)

a^x = e^x * lna (standard log rules)

since lna is a constant, this can now be integrated the same method as e^kx

so your answer is 1/lna * e^xlna

=1/lna * a^x

for those confused, you cannot differentiat this in the same was as x^a, since x is a variable rather than a constant

a^x = e^x * lna (standard log rules)

since lna is a constant, this can now be integrated the same method as e^kx

so your answer is 1/lna * e^xlna

=1/lna * a^x

for those confused, you cannot differentiat this in the same was as x^a, since x is a variable rather than a constant

#11

Isn't that what you'd do if it was x^a? You can't do that with a^x can you?

What's "a"? I assumed it was just another unknown variable.

#12

What's "a"? I assumed it was just another unknown variable.

We integrated with respect to a, not x

#13

a is normally an Arbitrary constant, meaning its just a number. x is normally a variable, therefore it changes.

if a was a variable, it would mean you would need to use the product rule, and find da/dx too

if a was a variable, it would mean you would need to use the product rule, and find da/dx too

#14

l3vity FTW!

Cheers mate, that's really useful.

Cheers mate, that's really useful.

#15

Ah well, this is why I study Accounting now and not Maths.

#16

haha anytime. i know how it feels to forget how to do a question like that.

#17

Isn't that what you'd do if it was x^a? You can't do that with a^x can you?

__No__

#18

Integrated it is (a^x+1)/x+1

This, then add a constant as well I think, so ( (a^(x+1)) / (x+1) ) + C

#19

This, then add a constant as well I think, so ( (a^(x+1)) / (x+1) ) + C

You're somewhat behind buddy