4.5 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 \(du\) is generally used to work with the chain rule. \[ u=g(x) \\ \int f(g(x))\cdot g´(x)\cdot dx = \int f(u)\cdot du \]
??? example To solve \(\int (x\sqrt{x-1})dx\): \[ let\ u=x-1 \\ ∴ \frac{du}{dx}=1 \\ ∴ du=dx \\ \begin{align*} \int (x\sqrt{x-1})dx &\to \int(u+1)(u^\frac{1}{2})du \\ &= \int(u^\frac{3}{2}+u^\frac{1}{2})du \\ &= \frac{2}{5}u^\frac{5}{2}+\frac{2}{3}u^\frac{3}{2}+C \\ &= \frac{2}{5}(x-1)^\frac{5}{2} + \frac{2}{3}(x-1)^\frac{3}{2} + C \end{align*} \]
Definite integration
To find a numerical value of the area under the curve in the bounded interval \([a,b]\), the definite integral can be taken. \[\int^b_a f(x)dx\]
\(a\) and \(b\) are known as the lower and upper limits of integration, respectively.
(Source;
Kognity)
Regions under the x-axis are treated as negative while those above are positive, cancelling each other out, so the definite integral finds something like the net area over an interval.
If \(f(x)\) is continuous at \([a,b]\) and \(F(x)\) is the anti-derivative, the definite integral is equal to: \[\int^b_a f(x)dx=F(x)\biggr]^b_a=F(b)-F(a)\]
As such, it can be evaluated manually by integrating the function and subtracting the two anti-derivatives.
!!! warning If \(u\)-substitution is used, the limits of integration must be adjusted accordingly.
To find the total area enclosed between the x-axis, \(x=a\), \(x=b\), and \(f(x)\), the function needs to be split at each x-intercept and the absolute value of each definite integral in those intervals summed. \[A=\int^b_a \big|f(x)\big| dx\]
Properties of definite integration
The following rules only apply while \(f(x)\) and \(g(x)\) are continuous in the interval \([a,b]\) and \(c\) is a constant.
\[ \begin{align*} &\int^a_a f(x)dx&=&&0 \\ &\int^b_a c\cdot dx&=&&c(b-a) \\ &\int^b_a f(x)dx&=&&-\int^b_a f(x)dx \\ &\int^c_a f(x)dx&=&&\int^b_a f(x)dx + \int^c_b f(x)dx \end{align*} \]
The constant multiple and sum rules still apply.