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Find area bounded by y-axis from 3 to y1, let this region be X.
Let the point where x=2 intersects x -axis be B
Find area of ABOy1
Find area of trapezium ACOy1
Area of shaded region= Area of ABOy1- Area ofACOy1- region X

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First sub y=1.5 into eqn of the curve to find X1 and X2. find the area bounded by the curve from X1 to X2 and using the area found subtract the area of the rectangle below by using rectangle formula, 1.5 x (X2-X1)

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Find Y1 at P and Y2 at Q. Find area bound by curve from Y2 to Y1. Find area bounded by line from Y2 to Y1. Subtract area bound by line from area bound by curve

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Find area bound by curve from origin to B and area bound by the line from origin to C. Using area bound by line, subtract area bound by curve

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Ken

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Ken

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Let the y-coordinates of P and Q be y1 and y2 respectively
Find the area bounded by line y=1.5 from y1 to y2. We call it region A
Find the area bounded by y=1+sin(x/3) from y1 to y2.
We call it region B
Shaded region= Region B- Region A

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Ken

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Let the y coordinate of P be y2, Q be y1
Find the area bounded by graph from 0 to y2. We call this region A
Find the area of triangle of height Py1
Find the area bounded by graph from y1 to 0. We call this region B
Area of shaded region= Region A +(Area of triangle-region B)

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Sub y equals 1.5 into the eqn of the curve to find the x coordinates of the two interception points
As u have the two coordinates, u can find the area of the rectangle directly underneath the shaded region.
Then u can find the area of the shaded region by subtracting the area of the rectangle directly under the shaded region from the integral of the eqn of the curve from the two x coordinates found

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Let the origin be O, x-coordinate of B be x1
Find area of triangle AOC
Find area bounded by graph from 0 to x1. We call this region A
Shaded region= Area of triangle AOC-region A

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Let y coordinate of P Be y1 and coordinate of Q be (x1,y2). U can find the area of the shaded region on the left hand side of the y axis by intergrating the eqn of the curve from y1 to 0 with respect to y. Than u can make a big triangle, of which the area of the triangle is half times (y1-y2) times x1. Than u can get the area of the shaded region on the right hand side of the y axis by subtracting the intergral of the eqn of the curve from 0 to y2 with respect to y from the area of the big triangle previously found. Than u can find the total area of the shaded region by adding the two areas together

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For Area A, integrate with respect to y axis via manipulation of eqn so that no need to find eqn of line RQ and other shaded areas are found with calculation of area of triangle

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let coordinates of p and q be x1,y1 and x2,y2 respectively
integrate 1+sin(x/3) with respect to x from x1 to x2 to find value (a)
area of shaded region = (a) - area of rectangle pqx2x1

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3) with respect to x from x1 to x2 to find value (a) area of shaded region = (a) - area of rectangle pqx2x1

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to find area A:
integrate x=y^2-3y with respect to y from 0 to y-coordinate of P

to find area B:
find x-intercept of x+y=3 and let that point be R
find area of triangle OPR

to find area C:
integrate x=y^2-3y with respect to y from 0 to y-coordinate of Q

area of shaded region = A + B + C

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find area of triangle AOC to be value (z)

find area under curve:
let  x-coordinate of B be x1
integrate curve AB with respect to x from O to x1

area of shaded region=
(z) - area under curve

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area of AOC = 1/2 x b x h
integrate curve AB
subtract integrated curve from area of AOC