Question

In the problems of this section, set up and evaluate the integrals by hand and check your results by computer. $$\int_{x=0}^{4} \int_{y=0}^{x / 2} y d y d x$$

Short answer

\( \frac{8}{3} \)

Step-by-step solution

Understand the problem

The given integral is a double integral, which needs to be evaluated by setting up the integral and calculating it by hand. The limits for the outer integral are from 0 to 4 (with respect to x) and for the inner integral are from 0 to \( \frac{x}{2} \) (with respect to y).

Set up the inner integral

Write the inner integral with respect to y: \[ \begin{aligned} \int_{0}^{\frac{x}{2}} y \, dy \end{aligned} \]

Evaluate the inner integral

Integrate \( y \) with respect to y: \[ \begin{aligned} \int_{0}^{\frac{x}{2}} y \, dy &= \left[ \frac{y^2}{2} \right]_{0}^{\frac{x}{2}} \ &= \frac{\frac{x^2}{4}}{2} - \frac{0^2}{2} \ &= \frac{x^2}{8} \end{aligned} \]

Set up the outer integral

Substitute the value from the inner integral into the outer integral and set it up: \[ \begin{aligned} \int_{0}^{4} \frac{x^2}{8} \, dx \end{aligned} \]

Evaluate the outer integral

Integrate \( \frac{x^2}{8} \) with respect to x: \[ \begin{aligned} \int_{0}^{4} \frac{x^2}{8} \, dx &= \frac{1}{8} \left[ \frac{x^3}{3} \right]_{0}^{4} \ &= \frac{1}{8} \left( \frac{4^3}{3} - \frac{0^3}{3} \right) \ &= \frac{1}{8} \left( \frac{64}{3} \right) \ &= \frac{8}{3} \end{aligned} \]

Verify the result using a computer

Using a computer to check the double integral \[ \int_{x=0}^{4} \int_{y=0}^{x / 2} y \, dy \, dx \ \] also gives \( \frac{8}{3} \). This confirms that the manual calculation is correct.

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with the concept of integration. Integration is essentially the reverse process of differentiation and is used to find areas, volumes, central points, and many useful things.
When you see an integral, it consists of an integrand (the function you want to integrate), differential elements, and limits of integration if it's a definite integral.
In this exercise, we are working with double integrals, which extend the concept of integration to functions of two variables. Doubling up on integrations allows us to compute the volume under a surface in a 3D space.

Multiple Integration

Multiple integration involves integrating a function with more than one variable.
In this case, we have a double integral, meaning we integrate over two variables, x and y.
The double integral in the exercise is set up with nested integrals, where the inner integral is evaluated first and the outer integral second.
This method allows for breaking down a complex problem into simpler steps.

Definite Integrals

Definite integrals provide the area under a curve between two specified points, giving a concrete number as a result.
In the exercise, the limits of integration for y are from 0 to \(\frac{x}{2}\) and for x from 0 to 4.
The process involves finding the antiderivative of the integrand within the given limits. For the given exercise, we first find the integral of y from 0 to \(\frac{x}{2}\) before proceeding to the x limits.
The result of a definite integral is a specific numerical value that represents the area under the curve within the given boundaries.

Manual Evaluation of Integrals

Manual evaluation of integrals requires a step-by-step approach to break down the integral and solve it piece by piece.
In this exercise, we start by evaluating the inner integral with respect to y: \(\begin{aligned} \int_{0}^{\frac{x}{2}} y \, dy \end{aligned}\).
After calculating the inner integral, we replace it in the outer integral: \(\begin{aligned} \int_{0}^{4} \frac{x^2}{8} \, dx \end{aligned}\).
Each step involves finding the antiderivative and then evaluating it within specified limits, ultimately leading to the final answer, which was confirmed using a computer.