QuestionJuly 14, 2025

Evaluate the integral. int _(3)^8(1)/(x-1)dx int _(3)^8(1)/(x-1)dxapprox square (Type an integer or decimal rounded to three decimal places as needed )

Evaluate the integral. int _(3)^8(1)/(x-1)dx int _(3)^8(1)/(x-1)dxapprox square (Type an integer or decimal rounded to three decimal places as needed )
Evaluate the integral. int _(3)^8(1)/(x-1)dx int _(3)^8(1)/(x-1)dxapprox square (Type an integer or decimal rounded to three decimal places as needed )

Solution
4.7(251 votes)

Answer

0.847 Explanation 1. Identify the Integral The integral is \int_{3}^{8} \frac{1}{x-1} \, dx. 2. Determine the Antiderivative The antiderivative of \frac{1}{x-1} is \ln|x-1|. 3. Evaluate the Definite Integral Use the Fundamental Theorem of Calculus: \int_{a}^{b} f(x) \, dx = F(b) - F(a) where F(x) is an antiderivative of f(x). Compute: \left[ \ln|x-1| \right]_{3}^{8} = \ln|8-1| - \ln|3-1|. 4. Simplify the Expression Calculate: \ln(7) - \ln(2) = \ln\left(\frac{7}{2}\right). 5. Approximate the Result Use a calculator to find \ln\left(\frac{7}{2}\right) \approx 0.847.

Explanation

1. Identify the Integral<br /> The integral is $\int_{3}^{8} \frac{1}{x-1} \, dx$.<br /><br />2. Determine the Antiderivative<br /> The antiderivative of $\frac{1}{x-1}$ is $\ln|x-1|$.<br /><br />3. Evaluate the Definite Integral<br /> Use the Fundamental Theorem of Calculus: $\int_{a}^{b} f(x) \, dx = F(b) - F(a)$ where $F(x)$ is an antiderivative of $f(x)$.<br /> Compute: $\left[ \ln|x-1| \right]_{3}^{8} = \ln|8-1| - \ln|3-1|$.<br /><br />4. Simplify the Expression<br /> Calculate: $\ln(7) - \ln(2) = \ln\left(\frac{7}{2}\right)$.<br /><br />5. Approximate the Result<br /> Use a calculator to find $\ln\left(\frac{7}{2}\right) \approx 0.847$.
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