Question

Evaluate the integrals. $$ \int_{x=0}^{4} \int_{y=0}^{x / 2} y d y d x $$

Short answer

The value of the integral is \( \frac{8}{3} \).

Step-by-step solution

Integrate with respect to y

Start by integrating the inner integral with respect to y. The inner integral is: oin_{y=0}^{x/2} y \, d ySince the integral of y with respect to y is \( \frac{y^2}{2} \), we need to evaluate \[ \left. \frac{y^2}{2} \right|_{0}^{x/2} \].

Evaluate the inner integral

Evaluate the integrated inner function at the bounds y = 0 and y = x/2. Plug in the upper limit: \( \frac{(x/2)^2}{2} = \frac{x^2}{8} \). Plug in the lower limit: \( \frac{0^2}{2}= 0 \). The result of the inner integral is: \( \frac{x^2}{8} - 0 = \frac{x^2}{8} \).

Integrate with respect to x

Now, integrate the result of the inner integral with respect to x. The outer integral is now: \int_{0}^{4} \frac{x^2}{8} \, d xTo integrate \( \frac{x^2}{8} \), factor out \( \frac{1}{8} \) and integrate \( x^2 \) getting: \int_{0}^{4} \frac{x^2}{8} \, d x = \frac{1}{8} \int_{0}^{4} x^2 \, d xSince the integral of \( x^2 \) is \( \frac{x^3}{3} \), evaluate \left. \frac{x^3}{3} \right|_{0}^{4}.

Evaluate the outer integral

Evaluate the integrated outer function at the bounds x = 0 and x = 4. Plug in the upper limit: \( \frac{1}{8} \cdot \frac{4^3}{3} = \frac{1}{8} \cdot \frac{64}{3} = \frac{64}{24} = \frac{8}{3} \). Plug in the lower limit: \( \frac{1}{8} \cdot \frac{0^3}{3} = 0 \). The result of the outer integral is \( \frac{8}{3} - 0 = \frac{8}{3} \).

Key concepts

Iterated Integrals

Iterated integrals are a way to compute double integrals by breaking them into a sequence of single integrals. This method involves integrating with respect to one variable first, holding the other variable constant, and then integrating the result with respect to the second variable.
In our exercise, we deal with the integral \(\textstyle \int_{x=0}^{4} \int_{y=0}^{x / 2} y \ d y \ d x\). Remember, the inner integral is solved first, and its result is then used in the outer integral.
This process of nesting integrals is essential when tackling more sophisticated double integral problems. Whether integrating with respect to \(x\) or \(y\) first largely depends on the given bounds of integration.

Integration Bounds

Integration bounds are the limits between which you integrate your functions. They determine the start and end points for your integration. For double integrals, bounds can be more complex as they may depend on the variable of the other integral.
In our exercise, the bounds for the inner integral are \(y = 0\) to \(y = \frac{x}{2}\), which means for every fixed \(x\), \(y\) ranges from \(0\) to \(\frac{x}{2}\). This variable-dependent bound indicates a region in the xy-plane. For the outer integral, the bounds are constant \(x = 0\) to \(x = 4\). The overall area defined by these bounds forms a triangular region on the xy-plane.
Figuring out the correct bounds is crucial and often the trickiest part of solving multiple integrals. Visualizing the bounded region on a graph can be particularly helpful.

Calculus

Calculus is the mathematical study of continuous change. Integration, one of its core concepts, allows you to calculate areas under curves, accumulated quantities, and more. Double integrals extend this idea to functions of two variables, enabling computations for areas and volumes in higher dimensions.
The fundamental steps of integration involve finding antiderivatives and evaluating them at the given bounds. For instance, in solving our example problem, we utilized the basic antiderivative of \(y\) which is \(\frac{y^2}{2}\). Knowing how to apply these antiderivatives properly is key to handling integral calculus.
Additionally, understanding how calculus connects to real-world applications further enhances your problem-solving skills. From physics to engineering, mastering these concepts allows you to handle various practical problems efficiently.