# How do you Use the trapezoidal rule with n=8 to approximate the integral int_0^pix^2*sin(x)dx?

Aug 9, 2014

The "trap rule" approximates the area by creating n trapezoids with their bases on the x-axis, top corners along the curve y = f(x), and then adding their areas together.

Here we evaluate the function f(x) = x^2 sin(x) at 9 points along the interval from 0 to π, to make 8 intervals of $\Delta$x = π/8 for the trapezoid bases. x = 0, π/8, π/4, . . . , 7π/8, and π, then plug all these into the function f(x), to get the heights of the sides of the trapezoids: f(0) = 0^2 sin(0) = 0, f(π/8) = (π/8)^2 sin(π/8), etc.

Now use the trapezoid area formula: ${A}_{t r a p} = b \cdot \frac{{h}_{1} + {h}_{2}}{2}$
In our case $\Delta A = \Delta x \cdot \frac{f \left({x}_{i - 1}\right) + f \left({x}_{i}\right)}{2}$
and when you add all n of these together you get

sum_(i=1)^n DeltaA = Deltax*{f(x_0)+2f(x_1)+…+2f(x_(n-1))+f(x_n)}

because the middle terms appear twice, the right side of one trapezoid being the left side of the next. For our example,

A = sum_(i=1)^8= π/8[f(0)+2*f(π/8)+…+2*f((7π)/8)+f(π)].

You get to evaluate each term to as much accuracy as you need to get your answer to the specified tolerance. Happy calculator plugging!