The integral of e is e right? So if you were to take the integral of 24+e^(5t) (acceleration), it would be 24t+e^(5t) (velocity)?
No. Don't forget to include the derivative of the exponent. Remember that [itex] \frac{d} {{dt}}e^{5t} = e^{5t} \cdot 5 [/itex] - Warren
Alright, thanks for the help. :) I see what you're saying there. However, that's when taking the derivative, and I don't understand how to apply that. If I were taking the integral of e^(5t)*5, the 5 would be a constant, so since the constant of just plain e^(5t) is 1.. how could the integral change? Does it become like Ce^(5t) after integrated? When you say I have to include the derivative, do you mean (24+e^u du)dt where u = 5t, du = 5. For some reason I'm leaning towards du = 5 dt, dt = 1/5 du, so its (1/5)*24 du + e^u * (1/5) du, which when integrated becomes (24t + e^(5t))/5. I'm obviously confused. Maybe you could please show how to solve it for me? Integral of e
Whoops! That's not right. Just because 5 is a constant doesn't mean 5 e^(5t) is a constant. It's not; it's still a function of t. If you agree that the derivative of e^(5t) is 5 e^(5t), then you also have to agree that the integral of 5 e^(5t) is e^(5t) + C. What I mean when I say "don't forget to include the derivative of 5t," I mean this: You cannot directly integrate e^(5t). What you'd have to do is multiply it by 5 inside the integral sign, and divide it by 5 outside the integral sign, like this: [itex] \int {e^{5t} dt = \frac{1} {5}} \int {5e^{5t} dt} [/itex] Thus, the integral of e^(5t) is (1/5) e^(5t) + C. We don't solve homework problems. We help you understand how to solve them. - Warren
A lot of your problem is just sloppy thinking. You started by saying "The integral of e is e right?" No, the "integral of e" is NOT e. The integral of e (with respect to x) is ex since e is just a constant. If you MEANT that the integral of e^{x} is e^{x} (plus a constant NOT times a constant as you say) say that!