The Fundamental Theorem of Calculus and definite integrals
The Fundamental Theorem of Calculus:
$$\int_a^b F'(x)\,dx=F(b)-F(a).$$ A new notation for writing this: $$\left.F(x)\right|_a^b\equiv F(b)-F(a).$$
for example:
$$\int_1^3 x^2\,dx = \left.\frac{x^3}{3}\right|_1^3=\frac{(3)^3}{3}-\frac{(1)^3}{3}=\frac 93-\frac 13=\frac 83\approx2.667.$$
Substitution and limits
Using the substitution, $w(t)=t^2$ and $dw=w'(t)\,dt=2t\,dt$, $$\begineq\int_1^3 2te^{t^2}\,dt=& \int_1^3e^{\color{red}{t^2}}\color{black}{}\color{blue}{(2t\,dt)} \color{black}{=} \int e^w\,dw=e^w\\ =&\left.e^{t^2}\right|_1^3=e^{3^2}-e^{1^2}= e^9-e.\endeq$$
There's an alternative way to deal with the limits... $$\begineq\int_{t=1}^{t=3} 2te^{t^2}=&\int_{w(1)}^{w(3)} e^w\,dw\\ =&\left.e^w\right|_{w(1)}^{w(3)}=\left.e^w\right|_{1^2}^{3^2} =e^9-e.\\ \endeq$$
Example exam question
If $\int_0^6 f(t)= 8$, then $\int_0^3f(2x)\,dx=...$
- 4
- 5
- 6
- 8
- 10
- 16
Handout answers
$\int_0^4 6x\,dx=6\int_0^4 x\,dx=6\cdot\left.\frac{x^2}{2}\right|_0^4=\frac 62(4^2-0^2)=3\cdot 16=$48.
$\int_1^4\frac{1}{\sqrt x}\,dx=\int_1^4x^{-1/2}\,dx=\left.\frac{x^{1/2}}{1/2}
\right|_1^4
=\frac{(\sqrt 4-\sqrt 1)}{1/2}=2*(2-1)=$2.
$\int_0^1e^{-0.2t}\,dt$. Use $w(t)=-0.2t$ and $dw=w'dt=-0.2dt$:
$$\begineq\int_0^1e^{-0.2t}\frac{1}{-0.2}(-0.2\,dt)=&-5\cdot\int_{-0.2*0}^{-0.2*1}e^w\,dw\\
=&-5\left.e^w\right|_0^{-0.2}=-5*(e^{-0.2}-e^0)\\
=&-5e^{-0.2}-(-5)=5-5e^{-0.2}.\endeq$$
if (! $homepage){
$stylesheet="/~paulmr/class/comments.css";
if (file_exists("/home/httpd/html/cment/comments.h")){
include "/home/httpd/html/cment/comments.h";
}
}
?>