Chapter 35: Definite Integrals

Definite integrals formalize the "area" idea.

Note - The Definite Integral of f(x) from x = a to x = b

abf(x)dx\int_{a}^{b}f(x)dx

b - Upper Bound

a - Lower Bound

x - Integrand

d - Variable of Integration

The Integral sign \int  is also related to the letter S.

abf(x)dx=limn\int_{a}^{b}f(x)dx = \lim_{n\rightarrow \infty } k=1nf(a+kban)(ban)\sum_{k=1}^{n}f(a+k\frac{b-a}{n})(\frac{b-a}{n}) Riemann sum with n equally sized intervals

Example

abπdx=limnk=1nπ(10n)\int_{a}^{b}\pi dx =\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\pi \cdot (\frac{1-0}{n})

=limnk=1nπ(1n)=limnπ(1n)n=\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\pi (\frac{1}{n})=\lim_{n\rightarrow \infty }\pi \cdot (\frac{1}{n})\cdot n

=π= \pi

Thus, the total area is π\pi

Fact - k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}


Example - 011x2dx=23\int_{0}^{1}1-x^{2}dx=\frac{2}{3}

011x2dx=limNk=1Nf(0+k10N)(10N)\int_{0}^{1}1-x^{2}dx=\lim_{N\rightarrow \infty }\sum_{k=1}^{N}f(0+k\frac{1-0}{N})\cdot (\frac{1-0}{N})

=limNk=1Nf(kN)(1N)=\lim_{N\rightarrow \infty }\sum_{k=1}^{N}f(\frac{k}{N})(\frac{1}{N})

=limNk=1N[1(kN)2](1N)=\lim_{N\rightarrow \infty }\sum_{k=1}^{N}\left [ 1-(\frac{k}{N})^{2} \right ](\frac{1}{N})

=limN1Nk=1N[1(kN)2]=limN1N(k=1N1k=1N(kN)2)=\lim_{N\rightarrow \infty }\frac{1}{N}\sum_{k=1}^{N}\left [ 1-(\frac{k}{N})^{2} \right ]=\lim_{N\rightarrow \infty }\frac{1}{N}(\sum_{k=1}^{N}1-\sum_{k=1}^{N}(\frac{k}{N})^{2})

=limN1N(N1N2k=1Nk2)=\lim_{N\rightarrow \infty }\frac{1}{N}(N-\frac{1}{N^{2}}\sum_{k=1}^{N}k^{2})

=limN1N(N1N2N(N+1)(2N+1)6)=\lim_{N\rightarrow \infty }\frac{1}{N}(N-\frac{1}{N^{2}}\frac{N(N+1)(2N+1)}{6})

=limN11N3(2N3+3N2+N6)=\lim_{N\rightarrow \infty }1-\frac{1}{N^{3}}(\frac{2N^{3}+3N^{2}+N}{6})

=126=113=23=1-\frac{2}{6}=1-\frac{1}{3}=\frac{2}{3}


Discuss - 0110xdx=5\int_{0}^{1}10xdx=5 using k=1nk=n(n+1)2\sum_{k=1}^{n}k=\frac{n(n+1)}{2}


Rules for Definite Integrals

abf(x)dx=baf(x)dx\int_{a}^{b}f(x)dx= -\int_{b}^{a}f(x)dx

aaf(x)dx=0\int_{a}^{a}f(x)dx= 0 (No area)

abkf(x)dx=kabf(x)dx\int_{a}^{b}kf(x)dx= k\int_{a}^{b}f(x)dx

abf(x)+g(x)dx=abf(x)dx+abg(x)dx\int_{a}^{b}f(x)+g(x)dx= \int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx

abf(x)dx+bcf(x)dx=acf(x)dx\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx= \int_{a}^{c}f(x)dx

If f(x)f(x) has a min and max on [a,b][a,b]

(minf)(ab)abf(x)dx(maxf)(ab)(min f)(a-b)\leq \int_{a}^{b}f(x)dx\leq (max f)(a-b)

If f(x)g(x)on[a,b]f(x)\leq g(x)on [a,b] then

abf(x)dxabg(x)dx\int_{a}^{b}f(x)dx\leq \int_{a}^{b}g(x)dx


Geometry and Area

Example - Use geometry to calculate abf(x)dx\int_{a}^{b}f(x)dx

where f(x)={10x1x1x2f(x)=\left\{\begin{matrix} 1 & 0\leq x \leq 1 \\ x & 1\leq x\leq 2 \end{matrix}\right.

02f(x)dx=11+11+1211\int_{0}^{2}f(x)dx=1\cdot 1+1\cdot 1+\frac{1}{2}\cdot 1\cdot 1

=2+12=52=2+\frac{1}{2}=\frac{5}{2}


Example - Calculate 0442x2dx\int_{0}^{4}\sqrt{4^{2}-x^{2}dx}

0442x2dx=14(π42)\int_{0}^{4}\sqrt{4^{2}-x^{2}dx}=\frac{1}{4}(\pi \cdot4^{2} )

=4π=4 \pi


Summary

abf(x)dx=\int_{a}^{b}f(x)dx=  Area under graph

=limNk=1Nf(a+kbaN)(baN)=\lim_{N\rightarrow \infty }\sum_{k=1}^{N}f(a+k\frac{b-a}{N})(\frac{b-a}{N})






Note Created by
Is this note helpful?
Give kudos to your peers!
00
Wanna make this note your own?
Fork this Note
23 Views