# How do you find the area y = f(x) = 4/x^2 , from 1 to 2 using ten approximating rectangles of equal widths and right endpoints?

Sep 24, 2015

See the explanation.

#### Explanation:

$y = f \left(x\right) = \frac{4}{x} ^ 2$ , on $\left[1 , 2\right]$ using $n = 10$ approximating rectangles of equal widths and right endpoints.

$\Delta x = \frac{b - a}{n} = \frac{2 - 1}{10} = 0.1$

The bases of the rectangles are the subintervals:

$\left[1 , 1.1\right] , \left[1.1 , 1.2\right] , \left[1.2 , 1.3\right] , . . . , \left[1.8 , 1.9\right] , \left[1.9 , 2\right]$

The right endpoints are :

$1.1 , 1.2 , 1.3 , . . . , 1.9 , 2$

Use these to find the height of the rectangles:

$f \left(1.1\right) = \frac{4}{1.1} ^ 2 , f \left(1.2\right) = \frac{4}{1.2} ^ 2 , . . . , f \left(1.9\right) = \frac{4}{1.9} ^ 2 , f \left(2\right) = \frac{4}{2} ^ 2$

So the areas of the rectangles are $f \left({x}_{i}\right) \Delta x$ for these various ${x}_{i}$ and the Riemann sum is:

$\frac{4}{1.1} ^ 2 \Delta x + \frac{4}{1.2} ^ 2 \Delta x + \cdot \cdot \cdot + \frac{4}{1.9} ^ 2 \Delta x + \frac{4}{2} ^ 2 \Delta x$

$= \left(\frac{4}{1.1} ^ 2 + \frac{4}{1.2} ^ 2 + \cdot \cdot \cdot + \frac{4}{1.9} ^ 2 + \frac{4}{2} ^ 2\right) \Delta x$

$= \left(\frac{4}{1.1} ^ 2 + \frac{4}{1.2} ^ 2 + \cdot \cdot \cdot + \frac{4}{1.9} ^ 2 + \frac{4}{2} ^ 2\right) \left(0.1\right)$

Now do the arithmetic.