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:
one
Step-by-step explanation:
3x-7=4+6+4x
3x=7+4+6+4x
3x=17+4x
3x-4x=17
-x=17
x=-17
Answer:
I think it would be 5(4b+7)
Step-by-step explanation:
I divided both terms by 5.
Answer:
T F T T.
Step-by-step explanation:
Let p and q be two statement
When the value of p is true and the value of q is true then the value of 
is true.
When the value of p is true and the value of q is false then the value of is true.
When the value p is false and the value of q is true then the value of is false.
When the value of p is false and the value of q is false then the value of is true.
Hence, the value of is false when q is true and p is false.
