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.
We have:
\(\dfrac{\delta y}{\delta x}=f(x)g(x)\)
We want that in terms of \(y\).
We know from the product rule of differentiation:
\(y=a(x)b(x)\)
Means that:
\(\dfrac{\delta y}{\delta x}=a'(x)b(x)+a(x)b'(x)\)
So let’s relabel \(f(x)\) as \(h'(x)\)
\(\delta\)
\(\dfrac{\delta y}{\delta x}=h'(x)g(x)\)
\(\dfrac{\delta y}{\delta x}+h(x)g'(x)=h'(x)g(x)+h(x)g'(x)\)
\(y+\int h(x)g'(x)=\int h'(x)g(x)+h(x)g'(x)\)
\(y+\int h(x)g'(x)=h(x)g(x)\)
\(y=h(x)g(x)-\int h(x)g'(x)\)
For example:
\(\dfrac{\delta y}{\delta x}=x.\cos(x)\)
\(f(x)=\cos(x)\)
\(g(x)=x\)
\(h(x)=\sin(x)\)
\(g'(x)=1\)
So:
\(y=x\int \cos(x) dx-\int \sin(x)dx\)
\(y=x\sin(x)-\cos(x)+c\)
Take function \(f(x)\). From the extreme value theorem we know that:
\(\exists m \in \mathbb{R} \exists M\in \mathbb{R}\forall x\in [a,b](m<f(x)<M)\)
From continuation we know that:
\(\int_a^{x_1}f(x)dx+\int_{x_1}^{x_1+\delta x}f(x)dx=\int_a^{x_1+\delta x}f(x)dx\)
\(\int_x^{x_1+\delta x}f(x)dx=\int_a^{x_1+\delta }f(x)dx-\int_a^{x_1 }f(x)dx\)
Indefinite integrals
For later? Haven’t defined trigonometry yet.
\(f(c)=f(a)+\int^c_a \dfrac{\delta }{\delta x}f(x) dx\)