Let g(x) = f(t) dt, where f is the function whose graph is shown. (a) Evaluate g
ID: 2833187 • Letter: L
Question
Let g(x) = f(t) dt, where f is the function whose graph is shown. (a) Evaluate g(x) for x=0,4,8,12,16,20,and 24 (b) Estimate g(28). (use the midpoint to get the most preside estimate. g(28)= (c) Where does g have a maximum and a minimum value? minimum x= maximum x= Let g(x) = f(t) dt, where f is the function whose graph is shown. (a) Evaluate g(0), g(4) ,g(8), g(12), and g(24). (b) On What interval is g increasing? (Enter your answer using interval notation.) (c) Where does g have a maximum value? x=Explanation / Answer
In simple terms , intgeration of a curve is just the area under it with sign. Since I dont know what software it is that is being used, i will solve it manually.
For left curve,
g(4) = area under triangle formed by curve till 4 = 1/2 *base* Height = 1/2 * 4 * 4 = 8
g(8) = g(4) - area under triangle between 4 to 8 = 0
g(12) = g(8) - 1 triangle = - 8
g(16) = g(12)+ 1 triangle = 0
g(20) = g(16)+1 triangle + 1 rectangle = 8 +16 = 24
g(24) = g(20) + 1 triangle + 2 rectangle = 24 + 8 + 2*(16) = 64
The line between x=24 and 28 looks confusing. I am assuming its straight
Area of triangle formed by line between 24 and 28 = 1/2 * 4 * 12 = 24
g(28) = g(24)+24 = 88.
Minimum = point of highest values = x = 28
Minimum = point of lowest value = x = 12
For curve on the right
g(0) = 0
g(4) = 1 rectangle = 8 * 4 = 32
g(8)= g(4) + 1 triangle + 1 rectangle = 32 +16 +32 = 80
g(12)=g(8)+ 1 triangle = 80 + 1/2 * 4*16 =112
g(24)= g(12) - 1 triangle - rectangle = 112 - (1/2 * 8 * 8) - 4*8 = 48
g is increasing between 0<x<12
g has a maximum at x= 12
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