Question

Evaluate the double integral. $$ \int_{0}^{\infty} \int_{0}^{\infty} e^{-(x+y) / 2} d y d x $$

Short answer

The value of the double integral is \(2\).

Step-by-step solution

Identify the type of integral

This is a double integral defined from \(0\) to \(\infty\). Each integral will have to be calculated separately, starting with the inner integral.

Evaluate the inner integral

The inner integral is: \(\int_{0}^{\infty} e^{-(x+y) / 2} d y\). As this integral is a function of \(y\), \(x\) is a constant. Apply the formula for calculating the integral of an exponential function which is, \(\int e^{-x} dx = -e^{-x} \), adding the limits of integral, from \(0\) to \( \infty\).

Solve the limit

Treat \(x\) as a constant and apply the limits to \(y\). This will give us \( -2 e^{-(x+y) / 2}\Big|_0^\infty \). Evaluate this, which will result in \( -2e^{-x}\). This is the result of the inner integral. Now, this expression is used for the second integral.

Compute the outer integral

The outer integral is \(\int_{0}^{\infty} -2e^{-x} dx\). Evaluate this integral using the same rules as in Step 2. After integrating, and applying the limits we have, \(-2(-1) = 2\)

Combine results

Combine these results. The value of the double integral is \(2\).