Question

Use the midpoint approximation method to find the area under the curve 𝑓(𝑥)=1−𝑥² over

the interval [−1,1].

Answer 1

The midpoint rule (also known as the midpoint approximation) uses the midpoints of a subinterval:

  

$\begin{aligned} &\int_{a}^{b} f(x) d x \\ &\approx \Delta x\left(f\left(\frac{x_{0}+x_{1}}{2}\right)+f\left(\frac{x_{1}+x_{2}}{2}\right)+f\left(\frac{x_{2}+x_{3}}{2}\right)+\cdots+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)\right. \\ &\left.+f\left(\frac{x_{n-1}+x_{n}}{2}\right)\right) \\ \end{aligned}$

 

where  $\Delta x=\frac{b-a}{n}$

We have that $f(x)=1-x^{2}, a=-1, b=1 ,$ and $n=4$

Therefore,  $\Delta x=\frac{1-(-1)}{4}=\frac{1}{2}$

Divide the interval $[-1,1]$  into  $n=4$  subintervals of the length  $\Delta x=\frac{1}{2}$  with the following endpoints:  $a=-1,-\frac{1}{2}, 0, \frac{1}{2}, 1=b$

Now, just evaluate the function at the midpoints of the subintervals.

  

$\begin{aligned} &f\left(\frac{x_{0}+x_{1}}{2}\right)=f\left(\frac{-1-\frac{1}{2}}{2}\right)=f\left(-\frac{3}{4}\right)=\frac{7}{16}=0.4375 \\ &f\left(\frac{x_{1}+x_{2}}{2}\right)=f\left(\frac{-\frac{1}{2}+0}{2}\right)=f\left(-\frac{1}{4}\right)=\frac{15}{16}=0.9375 \\ &f\left(\frac{x_{2}+x_{3}}{2}\right)=f\left(\frac{0+\frac{1}{2}}{2}\right)=f\left(\frac{1}{4}\right)=\frac{15}{16}=0.9375 \\ &f\left(\frac{x_{3}+x_{4}}{2}\right)=f\left(\frac{\frac{1}{2}+1}{2}\right)=f\left(\frac{3}{4}\right)=\frac{7}{16}=0.4375 \end{aligned}$

Finally, just sum up the above values and multiply by  $\Delta x=\frac{1}{2} :$

 

$\frac{1}{2}(0.4375+0.9375+0.9375+0.4375)=1.375$