Question integration (1 Viewer)

[Dumdida]

aye peepz my fwend is asking a question but i duno it coz i cant memba it soz if anyone knowz....; find the area enclosed between the curve y=2/(x-3)^2, the x axis and the lines x=0 and x=1, thanx in advance

 

[Estel]

integral is -2/(x-3) over limits 0 and 1
i.e A = 1 - (2/3) = 1/3

 

[Slidey]

aye peepz my fwend is asking a question but i duno it coz i cant memba it soz if anyone knowz....; find the area enclosed between the curve y=2/(x-3)^2, the x axis and the lines x=0 and x=1, thanx in advance

It's an area question, so you treat y as the integrand and the x-values it is enclosed between as the limits. That is:

Integral of 2/(x-3)^2 from 0 to 1
To integrate this you'll need to recall that:
The indefinite integral of (ax+b)^n = [(ax+b)^(n+1)]/[a(n+1)]
In this case, a=1, b=-3, n=-2.
The integral is: -2/(x-3)
Applying the terminals 0 to 1:
1-2/3=1/3