Question

Evaluate the iterated integral. $$ \int_{1}^{4} \int_{1}^{e^{2}} \int_{0}^{1 / x z} \ln z d y d z d x $$

Short answer

The answer to this iterated integral is \(3e^2\).

Step-by-step solution

Understanding the Order of Integration

The exercise gives us the following iterated integral \(\int_{1}^{4} \int_{1}^{e^{2}} \int_{0}^{1 / x z} \(\ln z\) dy dz dx\). The first thing to observe is that we integrate with respect to \(dy\) first, then \(dz\), and finally \(dx\). So, our first task is to integrate the innermost integral with respect to \(y\).

Evaluating the innermost integral

The innermost integral is \(\int_{0}^{1 / x z} \(\ln z\) dy\). Since \(\ln z\) does not depend on \(y\), this integral simplifies to \( \[\ln z . (1/xz -0)\] = ln\ (z)/zx\).

Evaluating the second integral

Now we need to integrate this result with respect to \(z\). So, we get \(\int_{1}^{e^{2}} ln(z)/x.z dz\). We can recognize that there is a definite integral whose value is well known to be \(z \ln z - z\), thus the evaluation of this integral between 1 and \(e^2\) gives us \((e^2 \ln e^2 - e^2) - (1 \ln 1 - 1) = e^2\).

Evaluating the outermost integral

Finally we have to integrate this value with respect to \(x\). So, we get \(\int_{1}^{4} e^2 dx\). Since \(e^2\) is a constant, this integral simplifies to \(e^2\ (4-1)\) which is just \(3e^2\).