# How do you use the trapezoidal rule to approximate integral of e^-3x^2 dx between [0,1]?

Sep 3, 2015

Divide the domain $\left[0 , 1\right]$ into smaller trapezoids; calculate the area of each trapezoid; and sum to find an approximation.
Approximate area: $0.5 {e}^{- 3} \approx 0.24894$

#### Explanation:

Since ${e}^{- 3}$ is a constant and
since $\int c \cdot f \left(x\right) \mathrm{dx} = c \int f \left(x\right) \mathrm{dx}$ for any constant $c$

I will find an approximate value for ${\int}_{0}^{1} {x}^{2} \mathrm{dx}$ using trapezoids
and then multiply the result by ${e}^{- 3}$

For demonstration purposes I will divide the domain into 4 trapezoidal areas, each with a width of $0.25$

Trapezoid 1
from $0$ to $0.25$
has an area $= \frac{{0.25}^{2} - {0}^{2}}{2} \cdot 0.25 = 0.03125$
$\textcolor{w h i t e}{\text{XXX}}$(i.e. average height times width)

Similarly
Trapezoid 2 (from $0.25$ to $0.5$)
$\textcolor{w h i t e}{\text{XXX}}$has an area of $0.09375$
Trapezoid 3 (from $0.5$ to $0.75$)
$\textcolor{w h i t e}{\text{XXX}}$has an area of $0.15625$
Trapezoid 4 (from $0.75$ to $1.0$)
$\textcolor{w h i t e}{\text{XXX}}$has an area of $0.21875$

Total area of all 4 trapezoids
$\textcolor{w h i t e}{\text{XXX}} = 0.03125 + 0.09375 + 0.15625 + 0.21875$
$\textcolor{w h i t e}{\text{XXX}} = 0.5$

So ${\int}_{0}^{1} {x}^{2} \mathrm{dx} \approx 0.5$

and since ${\int}_{0}^{1} {e}^{- 3} {x}^{2} \mathrm{dx} = {e}^{- 3} {\int}_{0}^{1} {x}^{2} \mathrm{dx}$

${\int}_{0}^{1} {e}^{- 3} {x}^{2} \mathrm{dx} \approx 0.5 {e}^{- 3}$

If you wish to carry this further, note that ${e}^{- 3} \approx 0.049787$ (according to my calculator)