5.3¶
The Fundamental Theorem of Calculus¶
If \(f\) is continuous on the interval \([a,b]\) and \(f(t)=F'(t)\) for some function \(F\), then \(\int_{a}^{b}{f(t)dt}=F(b)-F(a)\)
\(\int_{a}^{b}{F'(t)dt}\) | \(=F(b)-F(a)\) |
---|---|
(Signed area between rate of change function and \(t\)-axis from \(t=a\) to \(t=B\)) | Change in F from \(t=a\) to \(t=b\) |
Ex.¶
If \(F'(t)=f(t)=\)velocity of an object in miles per hour, what are the units of \(\int_{a}^{b}{f(t)dt}\) and \(F(b)-F(a)\)
\(\int_{a}^{b}{f(t)dt}\approx \sum_{i}{f(t_1)\Delta t}\) \(=(\frac{\text{miles}}{\text{hour}})(\text{hours})=\text{miles}\)
F is distance, so it would in miles as well. Or by FTC, F(b)_F9A)=int_a^b(f(t)dt) so theyre in same units
5.4¶
Theorems about Definite Integrals¶
Thm: Properties of Limits of Integration¶
If \(a\),\(b\),\(c\) are numbers, and \(f\) continuous, then 1) \(\int_{b}^{a}{f(x)dx}=-\int_{a}^{b}{f(x)dx}\) 2) \(\int_{a}^{c}{f(x)dx}+\int_{c}^{b}{f(x)dx}=\int_{a}^{b}{f(x)dx}\)
ie:¶
\(\int_{a}^{b}{f(x)dx}\approx \sum_{i}{f(x_i)\Delta x}=\sum_{i}{f(x+i)\frac{b-a}{n}}\) \(\int_{b}^{a}{f(x)dx}\approx \sum_{i}{f(x_i)\Delta x}=\sum_{i}{f(x+i)\frac{a-b}{n}}\)
\(f(x)\) is even if \(f(-x)=f(x)\) (symmetric with respect to \(y\)-axis)
ie:¶
\(\cos{(-x)}=\cos{x}\)
\(f(x)\) is odd if \(f(-x)=-f(x)\) (symmetric about the origin) Hokora: "rotationally 180° symmetric!"
ie:¶
\(\sin{(-x)}=-\sin{(x)}\)
If \(f\) is even, then \(\int_{-a}^{a}{f(x)dx}=2\int_{0}^{a}{f(x)dx}\) If f is odd, then \(\int_{-a}^{a}{f(x)dx}=0\)
ie:¶
Given \(\int_{0}^{1.25}{cos(x^2)dx}=0.98\) \(\int_{0}^{1}{cos(x^2)dx}=0.90\) Find a) \(\int_{1}^{1.25}{cos(x^2)dx}\)
\(\int_{1}^{1.25}{cos(x^2)dx}=\int_{0}^{1}{cos(x^2)dx}+\int_{0}^{1.25}{cos(x^2)dx}\) \(0.98=0.90+\int_{0}^{1.25}{cos(x^2)dx}\) \(0.08=\int_{0}^{1.25}{cos(x^2)dx}\)
b) \(\int_{-1}^{1}{cos(x^2)dx}=2\int_{0}^{1}{cos(x^2)dx}=2(0.90)=1.80\)
c) \(\int_{1.25}^{-1}{cos(x^2)dx}=-\int_{-1}^{1.25}{cos(x^2)dx}\) \(=-(\int_{-1}^{1}{cos(x^2)dx}+\int_{1}^{1.25}{cos(x^2)dx})\) \(=-(1.80)+0.08)=-1.88\)
Thm: Properties of Sums and constant multiples of the Integrand¶
Let \(f\), \(g\) be continuous, and \(c\) \(a\) constant. \(\int_{a}^{b}{f(x)\pm g(x)dx}=\int_{a}^{b}{f(x)dx}\pm\int_{a}^{b}{g(x)dx}\) \(\int_{a}^{b}{cf(x)dx}=c\int_{a}^{b}{f(x)dx}\)
Ex:¶
Find exact value of \(\int_{0}^{2}{1+3xdx}=\int_{0}^{2}{1dx}+\int_{0}^{2}{3xdx}\) \(=\int_{0}^{2}{1dx}+3\int_{0}^{2}{xdx}\) \(=(\text{Area of rectangle})+3(\text{Area of triangle})\) \(=(2)(1)+3\xcancel{(1/2)}\xcancel{(2)}(2)\) \(=2+6=8\)
Average Value of Function¶
Average value of \(f\) from \(a\) to \(b\): \(\frac{1}{b-a}\int_{a}^{b}{f(x)dx}\)
ex:¶
\(C(t)\) is daily cost for heating house in dollars per day, \(t\) measured in days. time \(t=0\) corresponds to Jan 1, 2015
Q: What is \(\int_{0}^{90}{C(t)dt}\)? This is total cost in $ to heat house for 1st 90 days of 2015.
Q: What is \(\frac{1}{90}\int_{0}^{90}{C(t)dt}\)? Average cost per day to heat home during 1st 90 days of 2015.