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

1 Answer
Sep 3, 2015

Answer:

Divide the domain #[0,1]# into smaller trapezoids; calculate the area of each trapezoid; and sum to find an approximation.
Approximate area: # 0.5e^(-3) ~~ 0.24894#

Explanation:

Since #e^(-3)# is a constant and
since #int c*f(x) dx = c int f(x) dx# for any constant #c#

I will find an approximate value for #int_0^1 x^2 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 #=(0.25^2-0^2)/2*0.25 = 0.03125#
#color(white)("XXX")#(i.e. average height times width)

Similarly
Trapezoid 2 (from #0.25# to #0.5#)
#color(white)("XXX")#has an area of #0.09375#
Trapezoid 3 (from #0.5# to #0.75#)
#color(white)("XXX")#has an area of #0.15625#
Trapezoid 4 (from #0.75# to #1.0#)
#color(white)("XXX")#has an area of #0.21875#

Total area of all 4 trapezoids
#color(white)("XXX")= 0.03125+0.09375+0.15625+0.21875#
#color(white)("XXX")=0.5#

So #int_0^1 x^2 dx ~~ 0.5#

and since #int_0^1e^(-3)x^2 dx = e^(-3)int_0^1x^2dx#

#int_0^1 e^(-3)x^2 dx ~~ 0.5e^(-3)#

If you wish to carry this further, note that #e^(-3) ~~0.049787# (according to my calculator)