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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.8.32

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₀² dx / √|x − 1|

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1
First, recognize that the integral involves the expression \(\sqrt{|x - 1|}\), which has an absolute value inside the square root. This means the behavior of the integrand changes at \(x = 1\), so we should split the integral at this point.
Rewrite the integral as the sum of two integrals: \(\int_0^1 \frac{dx}{\sqrt{1 - x}} + \int_1^2 \frac{dx}{\sqrt{x - 1}}\). Notice how the absolute value affects the expression inside the square root on each interval.
For the first integral, \(\int_0^1 \frac{dx}{\sqrt{1 - x}}\), use the substitution \(u = 1 - x\), which implies $du = -dx$. Adjust the limits accordingly and rewrite the integral in terms of \(u\).
For the second integral, \(\int_1^2 \frac{dx}{\sqrt{x - 1}}\), use the substitution \(v = x - 1\), which implies $dv = dx$. Adjust the limits accordingly and rewrite the integral in terms of \(v\).
Evaluate both integrals using the power rule for integrals: \(\int u^{n} du = \frac{u^{n+1}}{n+1} + C\) for \(n \neq -1\). After evaluating, sum the results to get the value of the original integral.

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