I'm still having a hard time imagining how the area under a function relates to the definite integral.
I can derive the fundamental formula, and know that the definite integral is to take the limit of the riemann sum, as n approaches infinity.
For example, let say f(t) = t^2
∫ 3 0 f(x)dx = lim n→∞ ∑ n k=1 f(xk)Δx
= lim n→∞ ∑ n k=1 (3k/n)^2 * (3/n)
= lim n→∞ 27/n^3 * 1/6 * n(n+1)(2n+1)
= lim n→∞ 9/2n^2 * (2n^2 + 3n + 1)
= lim n→∞ 9 + 27/2n + 9/2n^2
= 9 which is x^3/3 when x = 3
I guess the question is how the riemann sum somehow transforms into the antiderivative of the function?
I can derive the fundamental formula, and know that the definite integral is to take the limit of the riemann sum, as n approaches infinity.
For example, let say f(t) = t^2
∫ 3 0 f(x)dx = lim n→∞ ∑ n k=1 f(xk)Δx
= lim n→∞ ∑ n k=1 (3k/n)^2 * (3/n)
= lim n→∞ 27/n^3 * 1/6 * n(n+1)(2n+1)
= lim n→∞ 9/2n^2 * (2n^2 + 3n + 1)
= lim n→∞ 9 + 27/2n + 9/2n^2
= 9 which is x^3/3 when x = 3
I guess the question is how the riemann sum somehow transforms into the antiderivative of the function?
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Yeah, the fundamental theorem of calculus is not intuitive----it's like magic! I'm not kidding, how one gets from a geometric interpretation like area under a curve to finding and evaluating an anti-derivative seems too good to be true!
It makes some intuitive sense when thinking about distance and velocity for example. If you chop up a rate of change curve (velocity) into a bunch of tiny little pieces and treat the curve like a sequence of constants over very small intervals, then total distance should be the sum of all products rate x times (integral of df/dt times dt).
That this generalizes is rather fantastic. The key to the proof is the Mean Value Theorem. Suppose that F is an anti-derivative of f on the interval [a, b]. If you chop up [a,b] into little pieces (form a partition),
a = x_0 < x_1 < x_2 < ...< x_n = b,
and look at one little sub-interval such as [x_0, x_1], the mean value theorem says that
F(x_1) - F(x_0) = F '(c)(x_1 - x_0)
for some c between x_0 and x_1. This means that
F(x_i) - F(x_(i-1)) = f(c_i) Δx_i
for each sub-interval in your partition---note that I used here that F '(c) = f(c). Add these up. On the left, you just get F(b) - F(a). On the right, you get ∫ f(x) dx over [a, b].
The Mean Value Theorem is the glue between differentiation and integration.
There are other fantastic results in calculus---all the fundamental theorems are like this in my opinion. Take Complex Variables when you get the chance. Residue calculus is equally thrilling!
It makes some intuitive sense when thinking about distance and velocity for example. If you chop up a rate of change curve (velocity) into a bunch of tiny little pieces and treat the curve like a sequence of constants over very small intervals, then total distance should be the sum of all products rate x times (integral of df/dt times dt).
That this generalizes is rather fantastic. The key to the proof is the Mean Value Theorem. Suppose that F is an anti-derivative of f on the interval [a, b]. If you chop up [a,b] into little pieces (form a partition),
a = x_0 < x_1 < x_2 < ...< x_n = b,
and look at one little sub-interval such as [x_0, x_1], the mean value theorem says that
F(x_1) - F(x_0) = F '(c)(x_1 - x_0)
for some c between x_0 and x_1. This means that
F(x_i) - F(x_(i-1)) = f(c_i) Δx_i
for each sub-interval in your partition---note that I used here that F '(c) = f(c). Add these up. On the left, you just get F(b) - F(a). On the right, you get ∫ f(x) dx over [a, b].
The Mean Value Theorem is the glue between differentiation and integration.
There are other fantastic results in calculus---all the fundamental theorems are like this in my opinion. Take Complex Variables when you get the chance. Residue calculus is equally thrilling!