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:
Pretty sure it’s D but if it is not then I don’t know
Answer:
One unit to the right.
Step-by-step explanation:
I entered both equations into desmos and rootx - 1 is one unit to the right.
Desmos is a great tool for graphing
Answer:
14x-35
Step-by-step explanation:
Use distributive property so,
you multiply 2x and -5 by 7
2x*7=14x
-5*7=-35
You can use factors to solve. Determine all the factor pairs of 24, find the two that are two numbers apart.
1, 24 X
2, 12 X
3, 8 X
4, 6 YES!
Algebraic way to solve using Quadratics:
l = 2 + w
A = lw
A = (2 + w)w Substitute (2 + w) for l
24 = (2 + w)w Substitute 24 in for the area
24 = 2w + w^2 Distribute
w^2 + 2w - 24 = 0 Set equal to 0 (put in standard form)
(w + 6) (w - 4) = 0 Factor
w + 6 = 0 and w - 4 = 0 Set each factor equal to 0.
So w= -6 or w = 4 ... -6 makes no sense for a length! So the width must be 4 and the length will be 4 + 2, which is 6.
