Question

Reverse the order of integration in the following integrals. $$\int_{1 / 2}^{1} \int_{0}^{-\ln y} f(x, y) d x d y$$

Short answer

The new integral, after reversing the order of integration, is $\int_{0}^{\ln 2} \int_{1}^{e^{-x}} f(x, y) dy dx$.

Step-by-step solution

Identify the region of integration

First, let's find the region of integration in the xy-plane. We can do this by using the bounds of integration in the given integral. For \(y\), we can see that it ranges from \(1/2\) to \(1\). For \(x\), it ranges from \(0\) to \(-\ln y\). Since we have a logarithmic function, we need to find the equation for this curve to be able to reverse the order of integration. The curve is given by \(x = -\ln y\) or \(-\ln y = x\).

Rewrite the constraint in an appropriate form

Since we want to rewrite the integral with respect to \(dy\), it is better to have the constraint in terms of \(y\). So, we need to rewrite the curve as a function of \(x\). We can do this by expressing \(y\) in terms of \(x\). From \(x = -\ln y\), we can write: $$y = e^{-x}$$

Determine the new bounds for x and y

Now, find the intersection points of the curve \(y = e^{-x}\) with the vertical bounds \(y = 1/2\) and \(y = 1\). When \(y = 1\): $$1 = e^{-x} \Rightarrow x = 0$$ When \(y = 1/2\): $$\frac{1}{2} = e^{-x} \Rightarrow x = \ln 2$$ So we have \(x\) ranging from \(0\) to \(\ln 2\). For \(y\), it will range from its minimum value \(y=1\) at \(x=0\) to its maximum value \(y=1/2\) at \(x=-\ln y\) or \(x = \ln 2\).

Rewrite the integral

Now we rewrite the original integral with the reversed order of integration, using the new bounds for \(x\) and \(y\) determined in step 3. The new integral is: $$\int_{0}^{\ln 2} \int_{1}^{e^{-x}} f(x, y) dy dx$$