Asked by sara
evaluate the integral from 4 to 5.
(x^3 -3x^2 -9)/(x^3 -3x^2)
(x^3 -3x^2 -9)/(x^3 -3x^2)
Answers
Answered by
Steve
first off, I'd use partial fractions:
(x^3 -3x^2 -9)/(x^3 -3x^2)
= 1 + 1/x + 3/x^2 - 1/(x-3)
There are no discontinuities on [4,5], so it's straightforward. The integral is just
x + logx - 3/x - log(x-3)
so we get
(5 + log5 - 3/5 - log2)-(4 + log4 - 3/4 - log1)
= 23/20 + log(5/8)
(x^3 -3x^2 -9)/(x^3 -3x^2)
= 1 + 1/x + 3/x^2 - 1/(x-3)
There are no discontinuities on [4,5], so it's straightforward. The integral is just
x + logx - 3/x - log(x-3)
so we get
(5 + log5 - 3/5 - log2)-(4 + log4 - 3/4 - log1)
= 23/20 + log(5/8)
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