Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the iterated integrals in Problems 1-14. \(\int_{1}^{2} \int_{0}^{x-1} y d y d x\)

Step-by-Step Solution

Answer

The value of the iterated integral is \(\frac{1}{6}\).

Step 1: Identify the Order of Integration

First, recognize that we have a double integral. The order of integration is important for solving this problem. The given integral is \(\int_{1}^{2} \int_{0}^{x-1} y \, dy \, dx\). This means we first integrate with respect to \(y\), and then with respect to \(x\).

Step 2: Integrate with Respect to y

To solve the inner integral, \(\int_{0}^{x-1} y \, dy\), treat \(x\) as a constant since we are integrating with respect to \(y\).The antiderivative of \(y\) with respect to \(y\) is \(\frac{y^2}{2}\). Evaluate this from \(y = 0\) to \(y = x - 1\):\[ \left[ \frac{y^2}{2} \right]_0^{x-1} = \frac{(x-1)^2}{2} - \frac{0^2}{2} = \frac{(x-1)^2}{2}. \]

Step 3: Substitute the Result into the Outer Integral

Substitute \(\frac{(x-1)^2}{2}\) for the inner integral back into the outer integral:\[ \int_{1}^{2} \frac{(x-1)^2}{2} \, dx. \]

Step 4: Simplify and Integrate with Respect to x

Simplify the integrand by factoring out the constant \(\frac{1}{2}\):\[ \frac{1}{2} \int_{1}^{2} (x-1)^2 \, dx. \]To solve this integral, expand \((x-1)^2 = x^2 - 2x + 1\):\[ \frac{1}{2} \left( \int_{1}^{2} (x^2 - 2x + 1) \, dx \right). \]Integrate term-by-term:- The integral of \(x^2\) is \(\frac{x^3}{3}\).- The integral of \(-2x\) is \(-x^2\).- The integral of \(1\) is \(x\).Apply from \(x = 1\) to \(x = 2\):\[ \frac{1}{2} \left( \left[ \frac{x^3}{3} - x^2 + x \right]_{1}^{2} \right). \]

Step 5: Evaluate the Integral at the Limits

Evaluate \(\left[ \frac{x^3}{3} - x^2 + x \right]\) from x = 1 to x = 2:1. Calculate at \(x = 2\): \[ \frac{2^3}{3} - 2^2 + 2 = \frac{8}{3} - 4 + 2 = \frac{8}{3} - 2 = \frac{2}{3}. \]2. Calculate at \(x = 1\): \[ \frac{1^3}{3} - 1^2 + 1 = \frac{1}{3} - 1 + 1 = \frac{1}{3}. \]Subtract these results:\[ \frac{2}{3} - \frac{1}{3} = \frac{1}{3}. \]Include the \(\frac{1}{2}\) factor:\[ \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}. \]

Step 6: Final Result

The value of the iterated integral is \(\frac{1}{6}\).

Key Concepts

Double Integrals

Double integrals are a way to calculate the volume under a surface in a rectangular region. These are similar to single integrals but apply to two variables instead of one. The notation for a double integral is typically:

• \( \int\int f(x, y) \, dx \, dy \)

The area of integration is often a rectangle in the \(xy\) plane, bounded by two intervals. In our exercise, the double integral is expressed as \( \int_{1}^{2} \int_{0}^{x-1} y \, dy \, dx \), showing that we integrate over a defined region.
We perform such integrals by first integrating with respect to one variable and then the other, lending our approach its name—iterated integration. By evaluating double integrals, we can determine not just areas but "volumes" above complicated regions.

Order of Integration

The order of integration is crucial in solving double integrals. Let's break this down. In our exercise, the double integral is set up as \( \int_{1}^{2} \int_{0}^{x-1} y \, dy \, dx \) with the inner integral (\(\int_{0}^{x-1} y \, dy\)) to be evaluated first. This sequence denotes that we integrate over \(y\) initially, with the outer integral (\(\int_{1}^{2} \, dx\)) applied afterward.
In certain scenarios, changing the order of integration may ease the evaluation process. However, care must be taken to adjust integration limits accordingly. The problem at hand shows the order \(dy\) then \(dx\), which is suitable in this case due to the provided limits. Understanding and recognizing these orders can provide better insights into evaluating them correctly.

Integration with Respect to a Variable

Integration with respect to a variable refers to treating the other variable as a constant while integrating. In our example, the inner integral \( \int_{0}^{x-1} y \, dy \) requires us to treat \( x \) as a constant.

• We find the antiderivative of \( y \) to be \( \frac{y^2}{2} \), then evaluate it across the limits from \( y = 0 \) to \( y = x - 1 \).
• This process is straightforward: calculating \( \frac{(x-1)^2}{2} - 0^2 \), resulting in \( \frac{(x-1)^2}{2} \).

Through this process, we simplify the problem's complexity by addressing one dimension at a time, which is the beauty of iterated integration.

Evaluating Definite Integrals

Evaluating a definite integral involves calculating the accumulated area or volume over a specified interval. After finding the inner integral, we substitute the result, \( \frac{(x-1)^2}{2} \), into the outer integral \( \int_{1}^{2} \frac{(x-1)^2}{2} \, dx \).

• We simplify by bringing the constant out: \( \frac{1}{2} \int_{1}^{2} (x-1)^2 \, dx \).
• Then expand \( (x-1)^2 \) and integrate it term-by-term: \( x^2 \), \( -2x \), and \( 1 \).
• Solving these from \( x=1 \) to \( x=2 \), and applying resulting values to find the difference simplifies to \( \frac{1}{3} \).
• Finally, incorporating the initial \( \frac{1}{2} \) factor provides the final result, \( \frac{1}{6} \).

Each step of evaluating definite integrals involves careful substitution, expansion, and calculation within boundaries set by the limits. Mastering these calculations brings precision in evaluating even more complex integrals.