Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:
![\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac2n%5Cright%5D%2C%5Cleft%5B%5Cdfrac2n%2C%5Cdfrac4n%5Cright%5D%2C%5Cleft%5B%5Cdfrac4n%2C%5Cdfrac6n%5Cright%5D%2C%5Cldots%2C%5Cleft%5B%5Cdfrac%7B2%28n-1%29%7Dn%2C2%5Cright%5D)
Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,
where 
. Each interval has length 
.
At these sampling points, the function takes on values of
We approximate the integral with the Riemann sum:
Recall that
so that the sum reduces to
Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:
Just to check:
1 pound = 16 ounces
8 * 16 = 128 ounces
Answer:
34.875
Step-by-step explanation:
To round a number, you look to the next place to the right of what you want to round to... for example, if you want to round to the nearest hundredth, you look to the thousandths place to see whether the hundredths place rounds up or down.
0.04 is already at the hundredths place so the thousandths place is zero... the answer to the nearest hundredth is 0.04.
For 0.2%, you have to convert the percentage to a decimal by dividing by 100 (move the decimal 2 places to the left).
0.2% = 0.002
So now to round to the nearest hundredth, we look to the thousandths place. The thousandths place has a 2 in it so the hundredths place rounds down. 0.2% total he nearest hundredth is 0.
