Calculus
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/23=0 Or x^23x+2=0
Or (x1)(x2)=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}*(21)2log2=3/22log2, but answer shown is 3/2log2. Am I committing some mistake somewhere? Please advise.
asked by
MS

Looks good to me. I get
∫[1,2] (3x)  2/x dx
= 3x  1/2 x^2  2logx [1,2]
= 3/2  2log2
gotta be a typo in the answer. 2log2 is correct.posted by Steve

Thanks Mr. Steve for the guidance.
posted by MS
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