Question

In Exercises \(53-66,\) make a \(u\) -substitution and integrate from \(u(a)\) to \(u(b) .\) $$\int_{1}^{2} \frac{d t}{t-3}$$

Short answer

The solution to the integral \(\int_{1}^{2} \frac{d t}{t-3}\) is \( - \ln 2 \).

Step-by-step solution

Identify the Substitution

Identify \(u\) as the denominator in the integral function. So, set \(u = t - 3\).

Derive \(du\)

Now differentiate \(u = t - 3\) with respect to \(t\) to obtain \(du\). This yields \(du = dt\).

Update the Limits of the Integral

Substitute the given limits (\(1\) and \(2\)) into the expression for \(u\) to get the new limits of the integral. You have \(u(1) = 1 - 3 = -2\) and \(u(2) = 2 - 3 = -1\). So the new limits are \(u(1) = -2\) and \(u(2) = -1\).

Substitute \(u\) and \(du\) into the Integral

Now replace \(t - 3\) with \(u\) and \(dt\) with \(du\) in the integral. The integral becomes: \(\int_{-2}^{-1} \frac{du}{u}\).

Integrate

The integral of \(\frac{1}{u}\) with respect to \(u\) is \( \ln |u|\). Therefore, \(\int_{-2}^{-1} du/u = [\ln |u|]_{-2}^{-1}\). Interpret this as \(\ln |-1| - \ln |-2|\) .

Compute the Definite Integral

Plug in the values of \(u\) to get: \( \ln 1 - \ln 2 = - \ln 2\).