# Thread: Using Midpoint rule to appromiate area bounded by curvves

1. ## Using Midpoint rule to appromiate area bounded by curvves

I thought I had the process but I was off apparently. The equations are y=sin^2(pi x/4), y=cos^2(pi x/4) 0, greater or equal to x, greater or equal to 1)

2. Originally Posted by sfgiants13
I thought I had the process but I was off apparently. The equations are y=sin^2(pi x/4), y=cos^2(pi x/4) 0, greater or equal to x, greater or equal to 1)
The directions of the inequalities are wrong, but assume you mean for $0 \le x \le 1$. Then area is:

$A=\int_{x=0}^1 \cos^2(\pi x/4)-\sin^2(\pi x/4)\ dx$

now what is the difficulty?

CB

3. Originally Posted by CaptainBlack
The directions of the inequalities are wrong, but assume you mean for $0 \le x \le 1$. Then area is:

$A=\int_{x=0}^1 \cos^2(\pi x/4)-\sin^2(\pi x/4)\ dx$

now what is the difficulty?

CB
Yeah sorry...it was use the midpoint rule and n=4 to approximate the area between the curves.

4. Originally Posted by sfgiants13
Yeah sorry...it was use the midpoint rule and n=4 to approximate the area between the curves.
You want four subintervals of the interval interval $[0,1]$, so they are each of length $0.25$, and have mid points $0.125, 0.375, 0.625, 0.875$, or:

$x_i=\frac{b-a}{n} \times (k+0.5), \ \ k=0,1,.., n-1$

are the interval mid points, where $a, b$ are the limits of integration, and $n$ is the number of subintervals.

Then:

$\int_a^b f(x)\ dx \approx \frac{b-a}{n}\sum_{k=0}^{n-1}f(x_i)$

or writing $h=(b-a)/n$

$x_i=h \times (k+0.5), \ \ k=0,1,.., n-1$

$\int_a^b f(x)\ dx \approx h\sum_{k=0}^{n-1}f(x_i)$

which is the sum of the function at the mid points of the intervals times the interval length

CB