Evaluate the integral 15^x dx on -1,1

Question

anonymous · Answer

$$\int\limits_{-1}^{1}15^{x}dx$$

anonymous · Answer

15^x=e^(log(15^x))=e^(log(15)x)

australopithecus · Answer

So,

$$\int\limits{a^u}du = \frac{a^u}{\ln(a)}  + c$$

anonymous · Answer

I've gotten as far as $$\frac{ 15^x }{\ln15 }$$
But when I plug in -1 and 1 and subtract, my answer doesn't match.

australopithecus · Answer

Blah I forgot the notation but you need to subtract the lower limit from the upper limit

australopithecus · Answer

15/ln(15) - (15^-1)/ln(15)

australopithecus · Answer

Do you follow?

anonymous · Answer

Yes. That's what I was doing but the answer is supposed to be $$\frac{ 244 }{15\ln15}$$
I can see how the bottom turns to 15ln15 but the 244 has be completely lost.

australopithecus · Answer

remember this is fractional subtraction

australopithecus · Answer

matching denominators and all that jazz

australopithecus · Answer

that is the same answer as what I posted above as well

australopithecus · Answer

https://www.wolframalpha.com/input/?i=+15%2Fln%2815%29+-+%2815^-1%29%2Fln%2815%29

anonymous · Answer

Ah, of course! It all makes sense now! 
Thank you so much for pointing that out!!

australopithecus · Answer

No problem anytime :D I hope your life is going well

australopithecus · Answer

and I just got your name on here ugh I'm ditzy

australopithecus · Answer

https://www.wolframalpha.com/input/?i=integrate+between+-1+and+1+for+15^x+dx If you didnt know how to integrate using wolfram alpha

australopithecus · Answer

Sometimes the answer will look different from what you got but in reality it is the same so I really recommend using wolfram alpha to save yourself the time

anonymous · Answer

I was trying to use it but couldn't figure out how to enter the bounds. Now I know!