2.9 KiB
SL Math - Analysis and Approaches - 2
The course code for this page is MCV4U7.
Integration
Integration is an operation that finds the net area under a curve, and is the opposite operation of differentiation. As such, it is also known as anti-differentiation.
The area under a curve between the interval of x-values \([a,b]\) is: \[A=\lim_{x\to\infty}\sum^n_{i=1}f(x_i)\Delta x\]
which can be simplified to, where \(dx\) indicates that integration should be performed with respect to \(x\): \[A=\int^b_a f(x)dx\]
While \(\Sigma\) refers to a finite sum, \(\int\) refers to the sum of a limit.
As integration is the opposite operation of differentiation, they can cancel each other out. \[\frac{d}{dx}\int f(x)dx=f(x)\]
The integral or anti-derivative of a function is capitalised by convention. Where \(C\) is an unknown constant: \[\int f(x)dx=F(x)+C\]
When integrating, there is always an unknown constant \(C\) as there are infinitely many possible functions that have the same rate of change but have different vertical translations.
!!! definition - \(C\) is known as the constant of integration. - \(f(x)\) is the integrand.
Integration rules
\[ \begin{align*} &\int 1dx &= &&x+C \\ &\int (ax^n)dx, n≠-1 &=&&\frac{a}{n+1}x^{n+1} + C \\ &\int (x^{-1})dx&=&&\ln|x|+C \\ &\int (ax+b)^{-1}dx&=&&\frac{\ln|ax+b|}{a}+C \\ &\int (ae^{kx})dx &= &&\frac{a}{k}e^{kx} + C \\ &\int (\sin kx)dx &= &&\frac{-\cos kx}{k}+C \\ &\int (\cos kx)dx &= &&\frac{\sin kx}{k}+C \\ \end{align*} \]
Similar to differentiation, integration allows for constant multiples to be brought out and terms to be considered individually.
\[ \begin{align*} &\int k\cdot f(x)dx&=&&k\int f(x)dx \\ &\int[f(x)\pm g(x)]dx&=&&\int f(x)dx \pm \int g(x)dx \end{align*} \]
Indefinite integration
The indefinite integral of a function contains every possible anti-derivative — that is, it contains the constant of integration \(C\). \[\int f(x)dx=F(x)+C\]
Substitution rule
Similar to limit evaluation, the substitution of complex expressions involving \(x\) and \(dx\) with \(u\) and \(du\) can be done.
??? example To solve \(\int (x\sqrt{x-1})dx\): \[ let\ u=x-1 \\ ∴ \frac{du}{dx}=1 \\ ∴ du=dx \\ \begin{align*} \int (x\sqrt{x-1})dx &\to \int(u+1)(u^\frac{1}{2})du \\ &= \int(u^\frac{3}{2}+u^\frac{1}{2})du \\ &= \frac{2}{5}u^\frac{5}{2}+\frac{2}{3}u^\frac{3}{2}+C \\ &= \frac{2}{5}(x-1)^\frac{5}{2} + \frac{2}{3}(x-1)^\frac{3}{2} + C \end{align*} \]