Given a function \(f(x)\) and an interval \([a,b]\), we can divide \([a,b]\) into \(n\) sections and calculate:
\(\sum_{j=0}^{n(b-a)}f(a+\dfrac{j}{n})\)
This is the Riemann sum.
We take the limit of the Riemann sum as \(n\rightarrow \infty\)
\(\int_a^b f(x)dx:= \lim_{n\rightarrow \infty } \sum_{j=0}^{n(b-a)} f(a+ \dfrac{j}{n} )\)
\(\int_a^bf(x)+g(x)dx=\lim_{n\rightarrow \infty }\sum_{j=0}^{n(b-a)}f(a+\dfrac{j}{n})+g(a+\dfrac{j}{n})\)
\(\int_a^bf(x)+g(x)dx=\lim_{n\rightarrow \infty }\sum_{j=0}^{n(b-a)}f(a+\dfrac{j}{n})+\lim_{n\rightarrow \infty }\sum_{j=0}^{n(b-a)}g(a+\dfrac{j}{n})\)
\(\int_a^bf(x)+g(x)dx=\int_a^bf(x)dx +\int_a^bg(x)dx\)
\(\int_a^bf(x)dx+\int_b^cf(x)dx=\lim_{n\rightarrow \infty }\sum_{j=0}^{n(b-a)}f(a+\dfrac{j}{n})+\lim_{n\rightarrow \infty }\sum_{j=0}^{n(c-b)}f(b+\dfrac{j}{n})\)
\(\int_a^bf(x)dx+\int_b^cf(x)dx=\lim_{n\rightarrow \infty }[\sum_{j=0}^{n(b-a)}f(a+\dfrac{j}{n})+\sum_{j=0}^{n(c-b)}f(b+\dfrac{j}{n})]\)
\(\int_a^bf(x)dx+\int_b^cf(x)dx=\lim_{n\rightarrow \infty }[\sum_{j=0}^{n(b-a)}f(a+\dfrac{j}{n})+\sum_{j=n(b-a)}^{n(c-b)+n(b-a)}f(b+\dfrac{j-n(b-a)}{n})]\)
\(\int_a^bf(x)dx+\int_b^cf(x)dx=\lim_{n\rightarrow \infty }[\sum_{j=0}^{n(b-a)}f(a+\dfrac{j}{n})+\sum_{j=n(b-a)}^{n(c-a)}f(a+\dfrac{j}{n})]\)
\(\int_a^bf(x)dx+\int_b^cf(x)dx=\lim_{n\rightarrow \infty }[\sum_{j=0}^{n(c-a)}f(a+\dfrac{j}{n})]\)
\(\int_a^bf(x)dx+\int_b^cf(x)dx=\int_a^cf(x)dx\)
Definite integrals are between two points.
\(\int_0^1f(x)dx\)
Indefinite integrals are not. Eg +c at end. The antiderivative.
\(\int f(x)dx\)
\(\int_{[0,1]}f(x)dx\)
Taking the derivative of a function provides another function. The anti-derivative of a function is a function which, when differentiated, provides the original function.
As this function can include any additive constant, there are an infinite number of anti-derivatives for any function.