What is the area of the region between the graphs of f(x) = x^2 + 2xf(x) = x2 +2x, and g(x) = 2x+1g(x) = 2x+1g(x) = 2x + 1?

enclosed area A = 2/3 units^2

Step-by-step explanation:

Given:-

- The two functions are given:-

                  f ( x ) = x^2 + 2x

                  g ( x ) = 2x + 1

Find:-

What is the area of the region between the graphs of f(x) and g(x)

Solution:-

- We will determine the interval or limits in which the two given function f(x) and g(x).

- We will find the interval by determining the point of intersection between the two graphs. Hence,

                  f ( x ) = g ( x )

                  x^2 + 2x = 2x + 1

                  x^2 = 1

                  x = ± 1 , y = 3 , -1

- The pair of intersection coordinates are ( -1 , -1 ) and ( 1 , 3 ). The functions are bounded by the intervals:

                  -1 ≤ x ≤ 1

                  -1 ≤ y ≤ 3

- Now sketch each function to visualize the enclosed region. ( See attachment ). We see that in the range ( -1 ≤ x ≤ 1 ) the function g ( x ) > f ( x ).

- So we will find the area along the y-direction with the interval ( -1 ≤ x ≤ 1 ) will define the limits of integration. The enclosed area is given by:

- The limits of integration are  [ a , b ] = [ 1 , -1 ].

- The enclosed area A = 2/3 units^2.

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.