Question

Evaluate the double integral. $$ \int_{0}^{1} \int_{0}^{2}(x+y) d y d x $$

Short answer

The value of the given double integral is 3.

Step-by-step solution

Evaluate the Inner Integral

First, keep \(x\) as a constant and evaluate the inner integral with respect to \(y\): \(\int_{0}^{2}(x+y) d y\). This is simply a one-variable calculus problem. To solve this, the anti-derivative of \(x+y\) with respect to \(y\) is \(xy+y^2/2\). Evaluating this anti-derivative at the limits of integration (from 0 to 2) results in \(2x+2\).

Evaluate the Outer Integral

Next step is to evaluate the outer integral with respect to \(x\): \(\int_{0}^{1} (2x+2) d x\). Similar to the first step, find the anti-derivative of \(2x+2\) with respect to \(x\) which is \(x^2+2x\). Evaluating this anti-derivative at the limits of integration (from 0 to 1) gives a result of \(3\).

Key concepts

Inner Integral

When dealing with a double integral, the first step is often to focus on the inner integral. The inner integral is the part of the double integral where you integrate with respect to one variable while treating the other variable as a constant. In this exercise, the inner integral is \( \int_{0}^{2}(x+y) \, dy \). Here, the variable \( y \) is the one we are integrating over, and \( x \) is considered constant.

In single-variable calculus, calculating the integral of an expression like \( x+y \) with respect to \( y \) involves finding the anti-derivative, which is a function whose derivative brings you back to your original expression. For \( x+y \), the anti-derivative with respect to \( y \) is \( xy + \frac{y^2}{2} \).

To compute the definite integral, evaluate this anti-derivative at the upper and lower limits of \( y \), which are 2 and 0. Substitute \( y = 2 \) and \( y = 0 \) into \( xy + \frac{y^2}{2} \), and subtract the results to get \( 2x + 2 \). This simplification provides the reformulated expression to use in the outer integral.

Outer Integral

Once the inner integral is evaluated, the next step is to address the outer integral. The outer integral involves the remaining variable that was initially treated as a constant. In this case, it's the integral \( \int_{0}^{1} (2x+2) \, dx \). Here, you're integrating with respect to \( x \).

Similar to the inner integral process, the first step is to determine the anti-derivative of \( 2x + 2 \) in terms of \( x \). The anti-derivative is \( x^2 + 2x \). To find the definite integral, substitute the limits of integration, which are 1 and 0, into this expression.

Evaluate \( x^2 + 2x \) at \( x = 1 \) and \( x = 0 \), then subtract the results: \( (1^2 + 2 \times 1) - (0^2 + 2 \times 0) \). The calculation yields a value of 3. This result is the final value of the original double integral.

Anti-derivative

An anti-derivative, in the context of integration, is essentially the "reverse" of a derivative. It represents a function whose derivative is the given function we aim to integrate. Finding an anti-derivative is pivotal in solving integrals, particularly definite integrals such as in double integral calculations.

Given a function \( f(y) = x+y \), the process involves determining a function whose derivative with respect to \( y \) will return \( f(y) \). This function is \( g(y) = xy + \frac{y^2}{2} \), as adding a derivative of constant \( x \) with respect to \( y \) and the derivative of \( \frac{y^2}{2} \) yields the original function.

Similarly, for a function of \( f(x) = 2x+2 \), its anti-derivative with respect to \( x \) is \( g(x) = x^2+2x \). Understanding anti-derivatives is key to comprehending how to reverse the process of differentiation, thus allowing us to solve both definite and indefinite integrals across various applications.