Question

In the problems of this section, set up and evaluate the integrals by hand and check your results by computer. $$\int_{x=0}^{1} \int_{y=2}^{4} 3 x d y d x$$

Short answer

The value of the integral is 3.

Step-by-step solution

- Understand the Problem

We have a double integral \(\int_{x=0}^{1}\int_{y=2}^{4} 3x \ dy \ dx\) which means we need to integrate the function \(3x\) with respect to \(y\) first and then with respect to \(x\).

- Integrate with respect to y

Keep \(3x\) as a constant because we are integrating with respect to \(y\). The integration of \(3x \) with respect to \(y\) is \( 3xy \). Thus, \(\int_{y=2}^{4} 3x \ dy = 3x [y]_{2}^{4} = 3x(4 - 2) = 6x\).

- Integrate with respect to x

Now integrate \(6x\) with respect to \(x\) from 0 to 1. \(\int_{x=0}^{1} 6x \ dx\): \(\int_{x=0}^{1} 6x \ dx = 6 \int_{x=0}^{1} x \ dx\).

- Evaluate the integral

The integration \(\int x \ dx \) is \(\frac{x^2}{2}\), so: \(6 \int_{x=0}^{1} x \ dx = 6 [ \frac{x^2}{2} ]_{0}^{1} = 6 [ \frac{1^2}{2} - \frac{0^2}{2} ] = 6 [ \frac{1}{2} ] = 3\).

- Check with Calculation

You can use a calculator or a software like WolframAlpha: \(\int_{0}^{1} \int_{2}^{4} 3x dy dx = 3\).

Key concepts

Iterated Integrals

In calculus, double integration often involves iterated integrals. An iterated integral means performing integration one variable at a time. This process is essential when evaluating the area or volume under a surface. For instance, consider the double integral \(\false \int_{x=0}^{1}\int_{y=2}^{4} 3x dy dx\false \). Here, you first integrate with respect to y, treating other variables as constants. Once done, you integrate the resulting expression with respect to x. It's a step-by-step process that breaks down complex multi-variable functions into simpler single-variable integrations. Iterated integrals make solving double integrals more systematic and manageable.

Integration Techniques

Effective integration techniques are crucial when dealing with double integrals. For example, understanding how to handle constants during integration simplifies your calculation. In the problem \(\false \int_{x=0}^{1}\int_{y=2}^{4} 3x dy dx\false \), you see that the function inside is 3x. Here, during the first part of the integration, you treat 3x as a constant because you're integrating with respect to y. Post this integration, you end up with an easier expression to manage. Remember, if you integrate \(\false f(x)\ \times g(y)\false \) with respect to y first, only g(y) is affected while f(x) remains untouched. Techniques like substituting variables or recognizing patterns can also greatly assist in solving more advanced calculus problems involving integrals.

Calculus Problems

When tackling calculus problems, following a structured approach is imperative. Let's analyze our problem: \(\false \int_{x=0}^{1}\int_{y=2}^{4} 3x dy dx\false \). First, fully understand the given integral. Identify boundaries and the function you need to integrate. Next, perform the integration step-by-step: start with respect to y, then proceed with x. This precise methodology eliminates confusion and errors. Always double-check your results, possibly using computational tools. Calculus problems enhance critical thinking and problem-solving skills as they involve breaking down complex tasks into workable steps. Such problems predominantly appear in engineering, physics, and other applied sciences, making them practical and valuable in real-world scenarios.