verification, (I am pretty sure, but would prefer a second opinion).

Question

idku · Answer

Just like$$\int\limits_{ }^{ }e^{x}~dx=e^x+C$$so is,
$$\int\limits_{ }^{ }e^{x+a}~dx=e^{x+a}+C$$(a is a constant)

and in my case,
$$\int\limits_{ }^{ }e^{x+1}~dx=e^{x+1}+C$$

myininaya · Answer

$$e^{x+a}=e^x e^a$$

idku · Answer

I understand that I would be going like$$\int\limits_{ }^{ }e^{x+a}~dx=e^a \int\limits_{ }^{ }e^{x}~dx=e^a e^{x}+C=e^{x+a}+C$$

jim_thompson5910 · Answer

You can verify your answer by deriving the e^(x+1) + C and you'll get e^(x+1) back again. So the answer is confirmed.

idku · Answer

that is why I am asking if I am correct, I get it, but I am not certain

myininaya · Answer

you verified what you already thought

idku · Answer

So it is correct?

myininaya · Answer

or I guess you could do the derivative thing suggested by jim too

anonymous · Answer

$$
d \left(e^{ax+b}\right) = e^{ax+b} ~d(ax+b) = ae^{ax+b}~dx
$$If your original function was divided by $a$, then: $$
d \left(\frac{e^{ax+b}}{a}\right)=\frac{e^{ax+b}}{a} ~d(ax+b) = e^{ax+b}~dx
$$

myininaya · Answer

but looks great to me

jim_thompson5910 · Answer

Rule:

$$\Large \int (f'(x))dx = f(x) + C$$

ie take the integral of f ' (x), and you get f(x). So you can check your answer by deriving f(x) with respect to x.

idku · Answer

yes differentiation would give the initial function, as well. tnx, great,

I will make that a formula fr myself.

idku · Answer

tnx everyone