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:

Answer:
12 girls
Step-by-step explanation:
multiply 8 by 3/2 to get 12.
Answer:

Step-by-step explanation:

Answer:
10
Step-by-step explanation:
A= L²
50 = L² (do opposite operation to find L)
L = 7.07
x²= L²+ L²
x²= 7.07² + 7.07²
x² = 100 ( do opposite operation to make x the subject)
x = 10
Note: opposite operation of square is square root