Asked by binzo
find the area between the graph of function and x-axis over the given interval, if possible. f(x)= 16/(x-1)^2 for (-infinity, 0]
Answers
Answered by
drwls
You want the integral of 16/(x-1)^2 from x = -infinity to zero.
For the indefinite integral, let x-1 = u
The indefinite integral of 16/u^2 is
16*(-1/u) = -16/u = -16/(x-1)
The integral from -inf. to 0 is
(-16/-1) - 0 = 16.
For the indefinite integral, let x-1 = u
The indefinite integral of 16/u^2 is
16*(-1/u) = -16/u = -16/(x-1)
The integral from -inf. to 0 is
(-16/-1) - 0 = 16.
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