I dont understand how antiderviative of 2 sin (2x) could be -cos(2x)

[mugzieee]

i don't understand how the antiderviative of 2 sin (2x) could be -cos(2x), what happens to the 2?

also how can the antiderivative of 5e^(5t) be e^(5t)?

 

[ek]

i don't understand how the antiderviative of 2 sin (2x) could be -cos(2x), what happens to the 2?

also how can the antiderivative of 5e^(5t) be e^(5t)?

Well, what do you get when you take the derivative of -cos2x? You apply the chain rule or whatever rule it is and come up with 2sin2x. Just like the 2 "appears" when taking the derivative, it also "disappears" when you take the antiderivative.

As for the second one, Dx (e^x) = e^x, that's why the second one is true.

If in doubt concerning an antiderivative, just check by taking the derivative of the answer you found.

 

[wais]

The derivative of 2 sin(2x) is -4cos(2x), so the antiderivative would be -2sin(2x). The 2 is still there, it just gets multiplied by the derivative of 2x which is 2. As for the antiderivative of 5e^(5t), the derivative of e^(5t) is 5e^(5t), so the antiderivative would be 5e^(5t)/5, which simplifies to just e^(5t). Remember, the antiderivative is the inverse operation of the derivative, so any constants or coefficients will also be affected by the inverse operation.