They only have 1 common point
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:

Step-by-step explanation:
The correct answer is 5i
Answer:
<em>$195</em>
Step-by-step explanation:
15a + 12c
5 adults and 10 children means a = 5 and c = 10
15a + 12c = 15(5) + 12(10) = 75 + 120 = 195
Answer:
0.79 sec
Step-by-step explanation:
Given there is a tool at the top of the building which is dropped by a worker and it follows the following equation at every instant of time .

where 
We know that this height is measured from the base of the building which means that when the tool reaches the bottom of the building it has h = 0 feet.
Let this be done at time t
h(t) = 0



t = 0.79 sec
Therefore the total time taken by the tool to reach the bottom of the building is 0.79 sec.