g(x) =f(t) dtwhere f is the function whose graph is shown in the figure.y16124t8-1-41345-812-16(a) Estimate g(0), g(2), g(4), g(6), and g(8).g(0) =g(2) =g(4) =g(6) =g(8) = (c) Identify any extrema of g.g has aSelect---ofat x =(d) Sketch a rough graph of g.yy8072248-8- 166456-2448- 3240-4032-4824- 56- 64- 72168-80468yy163282416468-88-16468-24-8O-32-16F

Given graph is,To find : g(0), g(2), g(4), g(6), g(8). Given that g(x)=∫0xf(t)dt To find: g(0) The integral over a trivial interval is zero. So g(0)=∫00f(t)dt=0. To find: g(2). Geometrically the area under the graph from 0 to 2 is g(2). so g(2)=∫02f(t) dt is the area between the graph f and horizontal x axis. This area forms a triangle with base b=2 and height h=16. and Area of triangle=12base×height. Thus g(2)=∫02f(t)dt=12(2)(16)⇒g(2)=16 Now to find g(4). Geometrically the area under the graph from 0 to 4 is g(4). so g(4)=∫04f(t) dt is the area between the graph f and horizontal x axis. This area forms a triangle with base b=4 and height h=16. and Area of triangle=12base×height. Thus, g(4)=∫04f(t)dt=12×4×16=32⇒g(4)=32. To find g(6). Geometrically the area under the graph from 0 to 6 is g(6). so g(6)=∫06f(t) dt is the area between the graph f and horizontal x axis. Here from 4 to 5 this area forms a triangle with base b=1 and height h=16 and from 5 to 6 this area forms a rectangle with length=1 and breadth=16. Thus, area of triangle=12×1×16=8 and area of rectangle=(1)×16=16. g(6)=∫06f(t)dt=g(4)+area of triangle +area of rectangle. g(6)=∫06f(t)dt=32+8+16⇒g(6)=56 To find g(8). Geometrically the area under the graph from 0 to 8 is g(8). so g(8)=∫08f(t) dt is the area between the graph f and horizontal x axis. Here from 7 to 8 this area forms a triangle with base b=1 and height h=16 and from 6 to 7 this area forms a rectangle with length=1 and breadth=16. Thus, area of triangle=12×1×16=8 and area of rectangle=(1)×16=16. Thus, g(8)=g(6)+area of triangle +area of rectangleg(8)=56+8+16⇒g(8)=80.Therefore, g(0)=0, g(2)=16, g(4)=32, g(6)= 56, g(8)=80.