C4 Integration Help!

Just unsure on how you would integrate 2^t with respect to t. Any help would be appreciated, cheers!

Just unsure on how you would integrate 2^t with respect to t. Any help would be appreciated, cheers!

You have to realize that a^x=e^xlna
because if we say a^x=e^bx then taking ln's of both sides lna^x=bx which then equals xlna=bx then dividing by x, lna=b

so 2^t=e^tln2

Sorry, correct me if I'm wrong but I'm pretty sure there is no e involved?

Sorry, correct me if I'm wrong but I'm pretty sure there is no e involved?

Try making the integrand "look like" the derivative of something that you already know.

a^x=e^(x*ln(a))

here a=2

You can't really integrate it without the e involved, you can't use the rules you have for e^x on 2^x

$y = 2^t$

$\ln y = \ln 2^t = t\ln 2$

$\frac{1}{y} \times \frac{dy}{dt}$ = $\ln 2$

$\frac{dy}{dt}$ = $y \ln 2 = 2^t \ln 2$

I see no need for the exponential function here?

You've used ln, which is the inverse of the exponential function. There is more than 1 way to do it.

BTW you haven't integrated it yet, you've differentiated it.

Yeah I know I said in my original post to use the differential to find the integral (basically do the reverse of the differential)

Yeah but it's not e though, is it? Anyhow glad we got that sorted out

Yeah I know I said in my original post to use the differential to find the integral (basically do the reverse of the differential)

Yeah but it's not e though, is it? Anyhow glad we got that sorted out

I'm still a bit confused. I previously understand how differentiating 2^t works, integrating is the only problem I have :/

If you think about it, integrating is just the opposite of differentiating, so say you have y = 2^t, then dy/dt = 2^t ln2, so then how would you integrate 2^t?

if you integrated dy/dt = 2^t ln2, you'd get 2^t, as you can see what's different about the 2 is the ln2 part, to get from dy/dt to y, you divide by ln2, so to integrate y, you divide y by ln2

Cheers!

Make the substitution $u = 2^t$, then $du = 2^t ln(2) dt$.

Or remember that $2^t = e^{ln(2^t)}$

Once you have $2^t = e^{ln(2^t)}$

How would you go about integrating it? Could you do it step-by-step please.

Because I can see how you would do the question by inspection and comparing it to the derivate of 2^t, but not how you would go about doing it with the base e.

All my friends say it's easier to pur everything to the base e, but I don't know how to go about integrating that (I reckon I am missing something obvious)

Thanks.

how do you follow through with this approach?