How do you use the trapezoidal rule with n=6 to approximate the area between the curve #f(x)=x^2-9# from -3 to 3?

1 Answer
Aug 23, 2015

Answer:

The area (between the x-axis and) #f(x)=x^2-9# from #-3# to #3#
is approximately #-55# square units (the negative indicates the area is below the x-axis)

Explanation:

Given #f(x)=x^2-9#

#{: (color(white)("X")x, color(white)("XXX"), f(x)), (-3, color(white)("XXX"), color(white)("X")0), (-2, color(white)("XXX"),-5), (-1, color(white)("XXX"), -8), ( color(white)("X")0, color(white)("XXX"), -9), (color(white)("X") 1, color(white)("XXX"), -8), (color(white)("X") 2, color(white)("XXX"), -5), ( color(white)("X")3, color(white)("XXX"), color(white)("X")0) :}#

Since the width of each trapezoid is 1, the area will be equal to the average height.

#{: ("trapezoid", color(white)("XX")"from x =", color(white)("XX")"to x=", color(white)("XX")"average height", color(white)("XX")"area" ), (" ", color(white)("XX")-3, color(white)("XX")-2, color(white)("XX")(0+(-5))/2=-2.5, color(white)("XX")-2.5 ), (" ", color(white)("XX")-2, color(white)("XX")-1, color(white)("XX")((-5)+(-8))/2=-6.5, color(white)("XX")-6.5 ), (" ", color(white)("XX")-1, color(white)("XXX")0, color(white)("XX")((-8)+(-9))/2=-8.5, color(white)("XX")-8.5 ), (" ", color(white)("XXX")0, color(white)("XXX")1, color(white)("XX")((-9)+(-8))/2=-8.5, color(white)("XX")-8.5 ), (" ", color(white)("XXX")1, color(white)("XXX")2, color(white)("XX")((-8)+(-5))/2=-6.5, color(white)("XX")-6.5 ), (" ", color(white)("XXX")2, color(white)("XXX")3, color(white)("XX")((-5)+0)/2=-2.5, color(white)("XX")-2.5 ) :}#

Total area of trapezoids
#= (-2.5)+(-6.5)+(-8.5)+(-8.5)+(-6.5)+(-2.5)#

#= -55#