Find the area of the region enclosed by f(x) and the x axis for the given function over the specified u reevaluate f(x)=x^2+3x+4 x<3 f(x)=4x+10 x greater than or equal to 3 on -3,4

A = 68 unit^2

Step-by-step explanation:

Given:-

The piece-wise function f(x) is defined over an interval as follows:

                       f(x) =  { x^2+3x+4    , x < 3

                       f(x) =  { x^2+3x+4     , x≥3

                        Domain : [ -3 , 4 ]

Find:-

Find the area of the region enclosed by f(x) and the x axis

Solution:-

- The best way to tackle problems relating to piece-wise functions is to solve for each part individually and then combine the results.

- The first portion of function is valid over the interval [ -3 , 3 ]:      

- The area "A1" bounded by f(x) is given as:

Where,  The interval of the function { -3 , 3 ] = [ a , b ]:

- Similarly for the other portion of piece-wise function covering the interval [3 , 4] :

- The area "A2" bounded by f(x) is given as:

Where,  The interval of the function { 3 , 4 ] = [ a , b ]:

- The total area "A" bounded by the piece-wise function over the entire domain [ -3 , 4 ] is given:

                     A = A1 + A2

                     A = 42 + 26

                     A = 68 unit^2

answer:

at time x=0 we have 1000, at time x=1 we have 2000, at time x=2 we have 4000.  

so that's an initial value of 1000 and a ratio of 2,

y = 1000 (2^x)

hope this !