Asked by MS
Find the area cut off by x+y=3 from xy=2. I have proceeded as under:
y=x/2. Substituting this value we get x+x/2=3
Or x+x/2-3=0 Or x^2-3x+2=0
Or (x-1)(x-2)=0, hence x=1 and x=2 are the points of intersection of the curve xy=2 and the line x+y=3. Area under curve above X axis for cut off segment is = Int 2/x dx from 1 to 2.
=[2log2] from 1 to 2 = 2log2.
Required area =Area of trapezoid - area under curve={(2+1)/2}*(2-1)-2log2=3/2-2log2, but answer shown is 3/2-log2. Am I committing some mistake somewhere? Please advise.
y=x/2. Substituting this value we get x+x/2=3
Or x+x/2-3=0 Or x^2-3x+2=0
Or (x-1)(x-2)=0, hence x=1 and x=2 are the points of intersection of the curve xy=2 and the line x+y=3. Area under curve above X axis for cut off segment is = Int 2/x dx from 1 to 2.
=[2log2] from 1 to 2 = 2log2.
Required area =Area of trapezoid - area under curve={(2+1)/2}*(2-1)-2log2=3/2-2log2, but answer shown is 3/2-log2. Am I committing some mistake somewhere? Please advise.