I just labelled f(x) as the original function (the antiderivative) and f(x)' is the derivative function in the integral, sinxcosx. Was that your misunderstanding?
And actually, the reason I did not understand what the d and du variables were, is because we just went over their uses today in...
My final process included the final equation, f(pi) = 1/4-cos(2pi) + C = 13.4; consequently, C = 13.65. In order to check my work of the antiderivative, I applied the chain rule to this original function. Consequently, f(x)' = 1/4(sin2x)(2) => 1/2sin(2x) or sinxcosx, which refers back to the...
Thanks alot, that makes sense in that way, but I ended up figuring out that I needed to apply the chain rule to the remainder of my initial thought process. Again, thanks a lot for the mental push :-)
Yeh, thanks, just took me several seconds to figure out the implied steps it took to get there...
Is this right?:
1/2[int]sin(2x)dx = 1/2(-cos(2x)) + C => 13.4 = 1/2(-cos(2pi)) + C => 13.4 - 1/2(-1) = C => 13.9 = C
Problem: [int]cosx(sinx)dx
Given: x=pi; f(pi)=13.4
I am utterly confused on how to solve this integral. I am 99% positive (which is nothing in the math world) that I need to apply the product rule to all of this in order to find the antiderivative. However, no matter how I think of going about...