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:
x ≥ 12
Step-by-step explanation:
Step 1: subtract 2 from both sides
Now you have; -3/4 x ≤ -9
Step 2: divide both sides by -3/4
Now you have x ≥ 12
(when dividing an inequality by a negative you have to flip the sign)
Therefore the answer is x ≥ 12
Tan (x) = 0.8. To calculate the angle x, find the tan⁻¹ (0.8) = 38.7°
Answer:
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Step-by-step explanation:
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