When y = 6,

3/(2x - 3) = 6

2x - 3 = 3/6 = ½

2x = 3½

x = 1¾ or 1.75

The intersection of the curve and the horizontal line has coordinates (1.75,6)


You already know the coordinates of P are (3,1)

To find the shaded region, integrate the expression of the curve 3/(2x - 3) with respect to x, from x = 1.75 to x = 3.

Add this to the area of the rectangle bounded by the y-axis, line y = 6, x-axis and a vertical line from (1.75,6) to the x-axis.

(Its width is 1.75 units , from x = 0 (the y-axis) to x = 1.75 . Its length(or height) is 6 units)

Then subtract the area of the right angled triangle with vertices O, P and the point on the x-axis directly below P (draw a dotted line from P to the x-axis) to get the shaded region.

This right angled triangle has base of 3 units and height 1 unit.


In summary,


∫³₁.₇₅ 3/(2x - 3) dx + (1.75 × 6) - (½ × 3 × 1)

= [3/2 ln(2x - 3)]³₁.₇₅ + 10.5 - 1.5

= 3/2 ln(2×3 - 3) - 3/2 ln(2×1.75 - 3) + 9

= 3/2 ln3 - 3/2 ln0.5 + 9

= 3/2 (ln3 - ln0.5) + 9

= 3/2 ln(3/0.5) + 9

= 3/2 ln6 + 9

Alternatively, integrate with respect to y from y = 1(where P is) to y = 6 and add the area to the area of the right angled triangle with vertices O, P and the point (0,1) on the y-axis, which is horizontal to P. This triangle has base of 1 unit and height 3 units also.

(Turn the graph sideways and let the y-axis be horizontal if you have difficulty visualising)

y = 3/(2x - 3)

2x - 3 = 3/y

2x = 3/y + 3

x = 3/2y + 1.5
x = 3/2 (1/y) + 1.5


∫⁶₁ (3/2 (1/y) + 1.5) dy + ½ × 1 × 3

= [3/2 lny + 1.5y]⁶₁ + 1.5

= 3/2 ln6 + 1.5×6 - (3/2 ln1 + 1.5×1) + 1.5

= 3/2 ln6 + 9 - 1.5 + 1.5

= 3/2 ln6 + 9

(In case you were wondering where the 3/2 ln1 went, recall that ln1 = 0)

Side note:

You can also find the area of the triangle by integrating x/3 with respect to x from x = 0 to x = 3

This is the area under the line which is equal to the area of the triangle.

∫³₀ x/3 dx

= [½x²/3]³₀

= [x²/6]³₀

= 3²/6 - 0²/6

= 9/6
= 1.5

If integrating with respect to y, it will be from y = 0 to y = 1

y = x/3
x = 3y

∫¹₀ 3y dy

= [3/2 y²]¹₀

= 3/2 × 1² - 3/2 × 0²
= 3/2
= 1.5



Of course, it's faster to do ½ × b × h