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 $\int_0^3f(2x)\,dx=...$

  1. 4
  2. 5
  3. 6
  4. 8
  5. 10
  6. 16
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$$\int_0^3f(2x)\,dx=?$$ use the substitution $w=2x$. Then $dw=w'\,dx=2\,dx$. So... $$\begineq\int_0^3f(2x)\,dx=&\int_0^3f(2x)\frac 12\cdot2\,dx\\ =&\frac 12\int_0^3f(2x)\cdot 2dx=\frac 12\int_{w(0)}^{w(3)}f(w)\,dw\\ =&\frac 12\int_{2*0}^{2*3}f(w)\,dw=\frac{1}{2}\int_0^6 f(w)\,dw. \endeq $$ But, $\frac{1}{2}\int_0^6 f(w)\,dw$ is the same definite integral as $\int_0^6 f(t)\,dt$, just written in terms of a variable $w$ instead of $t$. So we conclude that... $$\int_0^3f(2x)\,dx=\frac 12\int_0^6f(x)\,dx = \frac 12 8=4.$$ The answer is (a) 4

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 $dw=w'dt=-0.2dt$:
$$\begineq\int_0^1e^{-0.2t}\frac{1}{-0.2}(-0.2\,dt)=&-5\cdot\int_{-0.2*0}^{-0.2*1}e^w\,dw\\ =&-5\left.e^w\right|_0^{-0.2}=-5*(e^{-0.2}-e^0)\\ =&-5e^{-0.2}-(-5)=5-5e^{-0.2}.\endeq$$