Find the area under the curve of y=1/(3x-2)^0.5 between the limits x=1 and x=2 and the line y=0

This question requires integration since the area under the curve is equal to the integral between these bounds. Initially let u=3x-2 and differentiate with respect to x so then du/dx = 3. Rearrange to dx =du/3 and substitute this and u into the original integral. Then change the limits by substituting in x=2 for the upper limit and x=1 for the lower limit into u=3x-2. The new limits are then 4 ( for the upper ) and 1 (for the lower ). The integral is now : (u^-0.5)/3 du between limits u=4 and u=1. Integrating this gives [(2u^0.5)/3] (by 'adding one to the power and dividing by this new power'). Substitute in the calculated limits and subtract the upper from the lower limit as shown: (2(4)^0.5)/3 =4/3, (2(1)^0.5)/3 = 2/3, 4/3 -2/3 = 2/3. This gives a final area of 2/3 square units.