Asked by Raj
Find the area above the curve starting from 0 to 1 on x-axis of y=1/(2x+1)^2
Answers
Answered by
Damon
integral dx/(ax+b)^2 = -1/[ a(ax+b) ]
so
integral = -1/ [ 2(2x+1)] = -(1/2)/(2x+1)
when 1
= -(1/2)(3) = -3/2
when 0
= -1/2
-3/2 + 1/2 = -1
so
integral = -1/ [ 2(2x+1)] = -(1/2)/(2x+1)
when 1
= -(1/2)(3) = -3/2
when 0
= -1/2
-3/2 + 1/2 = -1
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