The shaded region in the diagram is bounded by the three parabolas y^2=36-9x , y=1/4x(x-4) , y^2=6x ?

Question

find the area of the shaded region.

Answer 1

Drawing a graph will be very helpful to you. Unfortunately, from the information you gave (without the graph), it is not clear which region is meant, as there are several regions bounded by these three parabolas.

Assuming that it is the middle-ish region, to the right of the y axis and mostly above the x axis:

The intersection of y^2=36-9x and y^2=6x is at 6x=36-9x or x=36/15=12/5.

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From x=0 to x=12/5

Bottom: y = (1/4)x(x-4)

Top: y^2 = 6x

Area:  int(x=0 to x=12/5) (sqrt(6x) - (1/4)(x^2-4x)) dx

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From x=12/5 to 4

Bottom: y = (1/4)x(x-4)

Top: y^2=36-9x

Area: int(x=12/5 to x=4) (sqrt(36-9x) - (1/4)(x^2-4x))

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This is going to get a bit messy, so I will leave some of the algebra to you. However, the important indefinite integrals are

int sqrt(6x) dx = sqrt(6)*x^(3/2)/(3/2) = (2/3)*sqrt(6)*x^(3/2)

int (x^2-4x) dx = x^3/3 - 2x^2

int sqrt(36-9x) dx = (36-9x)^(3/2) /((3/2)*-9) = -(2/27)*(36-9x)^(3/2)

Plug in the limits and you are done!