Integration (1 Viewer)

[Miss Sunshine]

Find the area enclosed between the curve y=3/(x-2), the y-axis and the lines y=1 and y=3, using the trapezoidal rule with 4 subintervals.

 

[life92]

Find the area enclosed between the curve y=3/(x-2), the y-axis and the lines y=1 and y=3, using the trapezoidal rule with 4 subintervals.

Because you need to find the area between the curve and the y axis, you need to make the subject of the formula x.

So y = 3 / (x-2)
xy -2y = 3
xy = 3 + 2y
x = (3+2y) / y
= 2 + 3/y

Now since we need 4 sub intervals, it means we will be taking the values of y; 1, 1.5, 2, 2.5, 3.

Now S (1=>3) 2 + 3/y dy
(approx) = 0.5 / 2 [ 2 + 3/1 + 2 (2 + 3/1.5 +2 + 3/2 +2 + 3/2.5) + 2 + 3/2]
= 7.475

We can check this by integrating, and if we do we get ~ 7.2958... which means our answer is correct.