Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The value of the iterated integrals is \( \frac{7}{2} \).

Step 1: Set up the Iterated Integral

The given iterated integral is \( \int_{1}^{2} \int_{0}^{x^{2}} \frac{y^{2}}{x} \ dy \ dx \). Here, we first focus on the interior integral with respect to \( y \), while treating \( x \) as a constant.

Step 2: Evaluate the Inner Integral

To evaluate the inner integral \( \int_{0}^{x^{2}} \frac{y^{2}}{x} \ dy \), factor out the \( \frac{1}{x} \) term: \[\frac{1}{x} \int_{0}^{x^{2}} y^2 \, dy.\]Next, compute the integral \( \int_{0}^{x^{2}} y^2 \, dy \).The antiderivative of \( y^2 \) is \( \frac{y^3}{3} \). Substitute the limits:\[\int_{0}^{x^{2}} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{x^2} = \frac{(x^2)^3}{3} - \frac{0^3}{3} = \frac{x^6}{3}.\]Thus, the evaluated inner integral is:\[\frac{1}{x} \cdot \frac{x^6}{3} = \frac{x^5}{3}.\]

Step 3: Evaluate the Outer Integral

Now evaluate the outer integral \( \int_{1}^{2} \frac{x^5}{3} \, dx \). First, factor out the \( \frac{1}{3} \):\[\frac{1}{3} \int_{1}^{2} x^5 \, dx.\]The antiderivative of \( x^5 \) is \( \frac{x^6}{6} \). Substitute the limits:\[\frac{1}{3} \left[ \frac{x^6}{6} \right]_{1}^{2} = \frac{1}{3} \left( \frac{2^6}{6} - \frac{1^6}{6} \right).\]Simplify further:\( 2^6 = 64 \) so:\[\frac{64}{6} - \frac{1}{6} = \frac{63}{6} = \frac{21}{2}.\]Thus, multiply this result by \( \frac{1}{3} \) to obtain the final value:\[\frac{1}{3} \times \frac{21}{2} = \frac{21}{6} = \frac{7}{2}.\]

Step 4: Simplify the Result

Finally, simplify the result to confirm: \( \frac{7}{2} \) is the simplest form of the evaluated integral. Thus, the value of the iterated integrals is \( \frac{7}{2} \).

Key Concepts

Antiderivative

When dealing with integrals, the antiderivative plays a crucial role. It's the reverse process of differentiation — instead of finding the derivative of a function, you're reconstructing the function from its derivative. In the context of iterated integrals, as seen in our exercise, you need to find antiderivatives with respect to each variable separately.

For example, to compute the inner integral \( \int_{0}^{x^{2}} y^2 \, dy \), you look for the antiderivative of \( y^2 \), which is \( \frac{y^3}{3} \). This antiderivative shows how the function \( y^2 \) accumulates area under the curve as \( y \) goes from \( 0 \) to \( x^2 \).

• Antiderivative of \( y^2 \) is \( \frac{y^3}{3} \).
• Antiderivative of \( x^5 \) is \( \frac{x^6}{6} \).

Once you have the antiderivative, you can evaluate it at the boundaries provided by the limits of integration. This method leads you to the accumulation of volumes or areas in the iterated integral.

Limits of Integration

Limits of integration define the boundaries over which the integration is performed. They are essential for calculating definite integrals as they specify the interval for integration.

In the given exercise, the limits of integration are set as \(0\) to \(x^2\) for the inner integral and \(1\) to \(2\) for the outer integral. It's crucial to integrate first within the limits of the inner integral because it often depends on the outer variable, like \( x \).So, you first calculate:

• For the inner integral: from \( y=0 \) to \( y=x^2 \).
• For the outer integral: from \( x=1 \) to \( x=2 \).

Visualizing the integration, think of how you cover the area or volume by sweeping through the given bounds, starting from the inner to the outer integral. This sequential process is key to properly evaluating iterated integrals.

Inner Integral

The concept of the inner integral revolves around the first integration step in an iterated integral. It's the integral that you solve first, usually with respect to one variable while treating the others as constants.

In our specific exercise, the inner integral is \( \int_{0}^{x^{2}} \frac{y^2}{x} \, dy \). Since \( x \) is treated as a constant here, you can factor out \( \frac{1}{x} \), simplifying the task of integration. This leaves you with \( \frac{1}{x} \int_{0}^{x^2} y^2 \, dy \), which is straightforward to evaluate once you've identified the antiderivative of \( y^2 \).

• Simplify by factoring constants (like \( \frac{1}{x} \)).
• Find the antiderivative of \( y^2 \).
• Substitute the limits after finding the antiderivative.

By focusing model integration on the inner bounds first, you effectively reduce the problem, allowing for a simpler solution when subsequently carrying out the outer integral.

Outer Integral

The outer integral represents the stage after you've evaluated the inner integral. You will now perform integration with respect to the remaining variable. This integral often concludes the process of solving iterated integrals.

In our example, once the inner integral \( \frac{x^5}{3} \) is calculated, attention shifts to the outer integral \( \int_{1}^{2} \frac{x^5}{3} \, dx \). Typically, you'll first factor any constant out again, such as \( \frac{1}{3} \), leading to simpler expressions to integrate.

• Factor out constants from the integral (e.g., \( \frac{1}{3} \)).
• Complete the integration using the antiderivative found previously (e.g., \( \frac{x^6}{6} \) for \( x^5 \)).
• Apply the limits of integration to the antiderivative.

After substituting the limits into the antiderivative, you achieve your final result. Simplifying this expression yields the outcome of the iterated integrals.