Antiderivative of e^-x and cos(mx)

[sapiental]

antiderivative of e^-x

I get -e^-x

antiderivative of cos(mx) where m is a constant

I get $$\frac {1}{m} \sin x$$

Please let me know if my attempts are correct. Thanks a lot!

 

[quasar987]

If you can calculate an ordinary derivative, you can always check whether the anti derivative you found is correct. For instance,

$$\frac{d}{dx}(-e^{-x}) = (-1)(-e^{-x})=e^{-x}$$

So your antiderivative is correct. Actually, the most general antiderivative to e^-x is -e^-x + C, where C is a constant. Do you see why?

 

[sapiental]

because the constant C changes the y value by a certain amount moving the functuion vertically on the graph?

This is what I know of the top of my head and am not quite sure how it relates to your question though.

 

[quasar987]

An antiderivative of a function f(x) is a function F(x) such that F'(x) = f(x). Is that your definition of antiderivative also? If it is, then suppose you found just such an antiderivative F(x). Then F(x) + C is also an anti derivative of f(x) because

$$\frac{d}{dx}(F(x)+C) = F'(x) + \frac{d}{dx}C = F'(x) + 0 =f(x)$$