Question

sketch the region of integration. (select the correct graph.)
\\( \int _ { 7 } ^ { \ln 7 } \int _ { e ^ { x } } ^ { 6 } \frac { 1 } { \ln y } d y d x \\)
evaluate the iterated integral, switching the order of integration if necessary.

Explanation:

Step1: Analyze the limits of integration

The given integral is \(\int_{0}^{\ln 7} \int_{e^{x}}^{7} \frac{1}{\ln y} \, dy \, dx\). For the inner integral, \(y\) goes from \(e^{x}\) to \(7\), and \(x\) goes from \(0\) to \(\ln 7\). To switch the order of integration, we need to describe the region in terms of \(y\) first.

We know that \(y = e^{x}\) implies \(x=\ln y\). When \(x = 0\), \(y=e^{0}=1\), but wait, actually when \(x = 0\), \(y\) starts at \(e^{0}=1\)? Wait no, the original limits: \(x\) from \(0\) to \(\ln 7\), and for each \(x\), \(y\) from \(e^{x}\) to \(7\). So when \(x = 0\), \(y\) starts at \(e^{0}=1\)? Wait, no, the lower limit of \(y\) is \(e^{x}\), and when \(x = 0\), \(e^{x}=1\), when \(x=\ln 7\), \(e^{x}=7\). Wait, no, that can't be. Wait, maybe I made a mistake. Wait, the upper limit of \(y\) is \(7\), and the lower limit is \(e^{x}\). So when \(x\) increases from \(0\) to \(\ln 7\), \(e^{x}\) increases from \(1\) to \(7\). So the region is bounded by \(y = e^{x}\) (or \(x=\ln y\)), \(y = 7\), \(x = 0\), and \(x=\ln 7\). So to switch the order, we need to find the limits for \(y\) and \(x\) in terms of \(y\).

For \(y\), it goes from \(1\) to \(7\) (since when \(x = 0\), \(y = e^{0}=1\), and when \(x=\ln 7\), \(y = e^{\ln 7}=7\)). For each \(y\) in \([1,7]\), \(x\) goes from \(0\) to \(\ln y\) (because \(y\geq e^{x}\) implies \(x\leq \ln y\), and \(x\geq 0\)). Wait, no, wait the original integral: \(x\) from \(0\) to \(\ln 7\), \(y\) from \(e^{x}\) to \(7\). So the region is under \(y = 7\), above \(y = e^{x}\), left of \(x=\ln 7\), and right of \(x = 0\). So when we switch the order, \(y\) will go from \(1\) to \(7\) (since when \(x = 0\), \(y\) starts at \(e^{0}=1\), and when \(x=\ln 7\), \(y = 7\)), and for each \(y\), \(x\) goes from \(0\) to \(\ln y\) (because \(x\) is between \(0\) and \(\ln y\) such that \(y\geq e^{x}\) (i.e., \(x\leq \ln y\)) and \(x\geq 0\)).

Step2: Switch the order of integration

So the integral becomes \(\int_{1}^{7} \int_{0}^{\ln y} \frac{1}{\ln y} \, dx \, dy\). Now we can integrate with respect to \(x\) first.

Step3: Integrate with respect to \(x\)

The inner integral is \(\int_{0}^{\ln y} \frac{1}{\ln y} \, dx\). Since \(\frac{1}{\ln y}\) is constant with respect to \(x\), the integral is \(\frac{1}{\ln y} \cdot x \big|_{0}^{\ln y} = \frac{1}{\ln y} \cdot (\ln y - 0) = 1\).

Step4: Integrate with respect to \(y\)

Now we have \(\int_{1}^{7} 1 \, dy\). The integral of \(1\) with respect to \(y\) is \(y\), so we evaluate from \(1\) to \(7\): \(7 - 1 = 6\).

Answer:

\(6\)