- #1
JinM
- 66
- 0
Hi everyone,
I know that integration is the inverse process of differentiation, and that the definite integral is defined as:
[itex]\int_{a}^{b} f(x) dx = \lim_{n \to \infty} \sum^{n}_{i = 1} f(x_i) \Delta x[/itex]
assuming that the integrand is defined over the interval [a,b].
My question is: Why is it that the antiderivatives give the area under the curve? For example:
[itex]\int \frac{1}{1+x^2} dx = \arctan{x} + C[/itex]
How does arctan give the area under curve of the integrand between a and b?
Thanks and much appreciated,
JinM
I know that integration is the inverse process of differentiation, and that the definite integral is defined as:
[itex]\int_{a}^{b} f(x) dx = \lim_{n \to \infty} \sum^{n}_{i = 1} f(x_i) \Delta x[/itex]
assuming that the integrand is defined over the interval [a,b].
My question is: Why is it that the antiderivatives give the area under the curve? For example:
[itex]\int \frac{1}{1+x^2} dx = \arctan{x} + C[/itex]
How does arctan give the area under curve of the integrand between a and b?
Thanks and much appreciated,
JinM