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How do you evaluate the definite integral #int dx/(1-x)# from #[-2,0]#?

Answer 1

Cameron Alexander

# int_(-2)^0 \ 1/(1-x) \ dx ln 3 #

We seek:

# I = int_(-2)^0 \ 1/(1-x) \ dx #

Noting that the integrand is continuous or over the range of integration, we can apply a simply substituting, Let

# u-1-x => (du)/dx = -1 #

And we have a transformation of the integration limits:

# x = { (-2),(0) :} => u = { (3),(1) :} #

Thus we have:

# I = int_(3)^1 \ 1/u \ (-1) \ du #

# \ \ = int_(1)^3 \ 1/u \ du #

# \ \ = [lnu]_(1)^3 #

# \ \ = ln 3 - ln 1 #

# \ \ = ln 3 #

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Answer 2

Ezekiel Alexander

To evaluate the definite integral (\int_{-2}^{0} \frac{dx}{1-x}), we first need to find the antiderivative of (\frac{1}{1-x}). The antiderivative of (\frac{1}{1-x}) is (-\ln|1-x|). Then, we evaluate this antiderivative at the upper and lower limits of integration and subtract the lower limit's value from the upper limit's value:

[ \begin{aligned} \int_{-2}^{0} \frac{dx}{1-x} &= \left[-\ln|1-x|\right]_{-2}^{0} \ &= -\ln|1-0| - (-\ln|1-(-2)|) \ &= -\ln(1) + \ln(3) \ &= -\ln(3) \end{aligned} ]

So, the value of the definite integral is (-\ln(3)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.