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?

Aug 23, 2015

The area (between the x-axis and) $f \left(x\right) = {x}^{2} - 9$ from $- 3$ to $3$
is approximately $- 55$ square units (the negative indicates the area is below the x-axis)

Explanation:

Given $f \left(x\right) = {x}^{2} - 9$

{: (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
$= \left(- 2.5\right) + \left(- 6.5\right) + \left(- 8.5\right) + \left(- 8.5\right) + \left(- 6.5\right) + \left(- 2.5\right)$

$= - 55$