The function
represents a reflection of
across the y-axis ⇒ 3rd answer
Step-by-step explanation:
Let us revise the reflection across the axes
- If the function f(x) reflected across the x-axis, then its image is g(x) = - f(x)
- If the function f(x) reflected across the y-axis, then its image is g(x) = f(-x)
∵ 
∵ g(x) is the image of f(x) after reflection across the y-axis
- From the rule above reflection across the y-axis changes the sign of x
∴ 
∵ 
∵ 
∴ 
∴ 
The function
represents a reflection of
across the y-axis
Learn more:
You can learn more about reflection in brainly.com/question/5017530
#LearnwithBrainly
Answer:
Step-by-step explanation:
1) factor the denominators: 2b/(b-1)2 - 2/ (b-1)2
2) MAKE SURE YOU HAVE A COMMON DENOMINATOR
3) combine the numerators: 2b-2/(b-1)2
4) distribute 2 from numerator: 2(b-1)/(b-1)2
5) simplify; answer is 2/(b-1)
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:
