Why ∫ 0 to 1 of sin(x) isn't the same as calculating the indefinite integral at x=1?

Question

I thought of this after reading https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Geometric_meaning

It turns out that the answer is just the difference between the indefinite integral at x=1 and x=0, but why not the indefinite integral at x=1.

Answer 1

I think this is a problem of definitions. What makes an integral "indefinite" is the fact that you haven't given it bounds, such as the range [0,1] in this case which restrict its value to a specific result. When you do, it becomes definite and has a value that can be compared to other results.

In your example the indefinite integral ∫sin(x) dx is simply - cos(x) + C. However, this is not evaluated at any point and can't be equal to a definite integral of any bounds because it has no value at all. To make the integral have a definite value, it must be taken over a range of values, as the area under the curve at any given point is 0. That said, even if you were to evaluate the anti-derivative at x = 1, the result is -cos(1) = ~ -0.5403 which is not equal to the integral evaluated over [0,1] . To evaluate over this range you do -cos(1) - ( -cos(0)) = ~0.46 .

Hope this helped!