Question

For the following integral, find the approximate value of the integral with 4 subdivisions using midpoint, trapezoid, and Simpsons approximation. Evaluate all trig functions, leave your answers with radicals when needed.

Answer 1

`\textsf{Midpoint rule}: \quad \frac{2\pi}{\sqrt[3]{2}}`

`\textsf{Trapezium rule}: \quad \pi`

`\textsf{Simpson's rule}: \quad (4 \pi)/(3)`

Explanation:

Midpoint rule

`\displaystyle \int_(a)^(b) f(x) \:\:\text{d}x \approx h\left[f(x_{(1)/(2)})+f(x_{(3)/(2)})+...+f(x_{n-(3)/(2)})+f(x_{n-(1)/(2)})\right]\\\\ \quad \textsf{where }h=(b-a)/(n)`

Trapezium rule

`\displaystyle \int_(a)^(b) y\: \:\text{d}x \approx (1)/(2)h\left[(y_0+y_n)+2(y_1+y_2+...+y_(n-1))\right] \quad \textsf{where }h=(b-a)/(n)`

Simpson's rule

`\displaystyle \int_(a)^(b) y \:\:\text{d}x \approx (1)/(3)h\left(y_0+4y_1+2y_2+4y_3+2y_4+...+2y_(n-2)+4y_(n-1)+y_n\right)\\\\ \quad \textsf{where }h=(b-a)/(n)`

Given definite integral:

`\displaystyle \int^(2 \pi)_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x`

Therefore:

• a = 0
• b = 2π

Calculate the subdivisions:

`\implies h=(2 \pi - 0)/(4)=(1)/(2)\pi`

Midpoint rule

Sub-intervals are:

`\left[0, (1)/(2)\pi \right], \left[(1)/(2)\pi, \pi \right], \left[\pi , (3)/(2)\pi \right], \left[(3)/(2)\pi, 2 \pi \right]`

The midpoints of these sub-intervals are:

`(1)/(4) \pi, (3)/(4) \pi, (5)/(4) \pi, (7)/(4) \pi`

Therefore:

`\begin{aligned}\displaystyle \int^(2 \pi)_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx (1)/(2)\pi \left[f \left((1)/(4) \pi \right)+f \left((3)/(4) \pi \right)+f \left((5)/(4) \pi \right)+f \left((7)/(4) \pi \right)\right]\\\\& = (1)/(2)\pi \left[\sqrt[3]{(1)/(2)} +\sqrt[3]{(1)/(2)}+\sqrt[3]{(1)/(2)}+\sqrt[3]{(1)/(2)}\right]\\\\ & = \frac{2\pi}{\sqrt[3]{2}}\\\\& = 4.986967483...\end{aligned}`

Trapezium rule

`\begin{array} c \cline{1-6} &&&&&\\ x & 0 & (1)/(2)\pi & \pi & (3)/(2) \pi & 2 \pi \\ &&&&&\\\cline{1-6} &&&&& \\y & 0 & 1 & 0 & 1 & 0\\ &&&&&\\\cline{1-6}\end{array}`

`\begin{aligned}\displaystyle \int^(2 \pi)_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx (1)/(2) \cdot (1)/(2) \pi \left[(0+0)+2(1+0+1)\right]\\\\& = (1)/(4) \pi \left[4\right]\\\\& = \pi\end{aligned}`

Simpson's rule

`\begin{aligned}\displaystyle \int^(2 \pi)_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx (1)/(3)\cdot (1)/(2) \pi \left(0+4(1)+2(0)+4(1)+0\right)\\\\& = (1)/(3)\cdot (1)/(2) \pi \left(8\right)\\\\& = (4)/(3) \pi\end{aligned}`