Question

Evaluate each of the iterated integrals. $$ \int_{0}^{3} \int_{0}^{1} 2 x \sqrt{x^{2}+y} d x d y $$

Short answer

The value of the iterated integral is \(-\frac{848}{15}\).

Step-by-step solution

Set Up the Inner Integral

The given problem is an iterated integral where we first integrate with respect to \(x\) and then with respect to \(y\). The inner integral we need to solve first is \(\int_{0}^{1} 2x \sqrt{x^2 + y} \, dx\).

Solve the Inner Integral

To integrate \( \int_{0}^{1} 2x \sqrt{x^2 + y} \, dx \), use the substitution \( u = x^2 + y \), which gives \( du = 2x \, dx \). Change the limits of \(x\) to \(u\): when \(x = 0\), \(u = y\) and when \(x = 1\), \(u = 1^2 + y = y + 1\). Now the integral becomes \( \int_{y}^{y+1} \sqrt{u} \, du \).

Solve the Substituted Integral

The integral \( \int_{y}^{y+1} \sqrt{u} \, du \) is equivalent to \( \int_{y}^{y+1} u^{1/2} \, du \). Use the power rule to integrate: \( \int u^{n} \, du = \frac{u^{n+1}}{n+1} \). Applying this, we get \( \frac{u^{3/2}}{3/2} \) evaluated from \(y\) to \(y+1\). Simplifying the expression, we find \( \frac{2}{3} \left[u^{3/2}\right]_{y}^{y+1} = \frac{2}{3} \left((y+1)^{3/2} - y^{3/2}\right) \).

Set Up the Outer Integral

Now the result of the inner integration serves as the new integrand for the outer integral with respect to \(y\): \( \int_{0}^{3} \frac{2}{3} \left((y+1)^{3/2} - y^{3/2} \right) \, dy \).

Solve the Outer Integral

The outer integral is \( \int_{0}^{3} \frac{2}{3} \left((y+1)^{3/2} - y^{3/2} \right) \, dy \). Distribute \(\frac{2}{3}\) and solve each term individually: \( \frac{2}{3} \int_{0}^{3} (y+1)^{3/2} \, dy - \frac{2}{3} \int_{0}^{3} y^{3/2} \, dy \). Use the power rule for each integral. The first integral evaluates to \( \frac{2}{3} \left[ \frac{2}{5} (y+1)^{5/2} \right]_{0}^{3} \) and the second integral evaluates to \( \frac{2}{3} \left[ \frac{2}{5} y^{5/2} \right]_{0}^{3} \).

Compute Definite Integrals and Subtract

Calculate \( \frac{2}{3} \times \frac{2}{5} \left[ (4)^{5/2} - (1)^{5/2} \right] \) for the first term, which evaluates to \( \frac{4}{15} (32 - 1) = \frac{124}{15} \). For the second term, calculate \( \frac{2}{3} \times \frac{2}{5} (27)^{5/2} \) for the second term, which evaluates to \( \frac{4}{15} (243) = \frac{972}{15} \). Subtracting these results yields \( \frac{124}{15} - \frac{972}{15} = -\frac{848}{15} \).

Simplify the Result

Simplify the subtraction to get the final result \(-\frac{848}{15}\).

Key concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, which allow us to calculate the area under a curve for a specific interval. In the context of iterated integrals, a definite integral is evaluated over a specific range or bounds, which are provided for each variable. For the exercise given, the iterated integrals have bounds [0,1] for the variable \(x\) and [0,3] for the variable \(y\).

• The 'bounds' refer to the limits between which the function will be integrated, and they are specified once for each variable.
• Iterated integrals often come in double integrals or higher, which means we integrate over multiple variables successively.
• The 'definite' nature of these integrals implies that we're looking for a numerical value that represents the total accumulation of our function within the specified limits.

In solving the problem, correctly identifying and applying the bounds is crucial, as it determines the scope of the integration process. After evaluating the definite integrals, we obtain a quantified value that represents the result of the iterated integration, expressed here as \(-\frac{848}{15}\). Understanding the role of definite integrals is essential for evaluating areas and cumulative quantities within specified intervals.

Integration Techniques

In calculus, integration techniques are essential tools for evaluating integrals when basic methods don't suffice. For iterated integrals, a combination of strategies is often used. In our exercise, we follow standard practices to handle the integration of complex functions.

• The inner integral is tackled by setting up the function \(2x \sqrt{x^2 + y}\) to be integrated with respect to \(x\), and substitution is applied to simplify the integrand.
• Once the inner integral is solved, the resultant expression is integrated over the outer variable \(y\).
• Integration techniques like substitution or using the power rule are common approaches for simplifying expressions and making integration feasible.

In particular, the power rule plays a significant role: it allows us to convert expressions with powers into integrals that can be solved by straightforward formulas. Effective integration requires patience and a consistent step-by-step approach to ensure all calculations and transformations are handled accurately.

Substitution Method

The substitution method is a widely-used integration technique often called "u-substitution". It is particularly useful for simplifying the integration of complex functions by reducing them to simpler forms.

• In our example, we start by identifying the expression \(u = x^2 + y\) to serve as our 'substitution'. This transforms a challenging integrand into a more manageable form.
• The derivative \(du = 2x \, dx\) matches a portion of the integrand, making the transformation valid and allowing us to rewrite the integral in terms of \(u\).
• The limits of integration also change from \(x\) to \(u\), by substituting the original bounds into the new variable expression.

This method simplifies the original difficult integral into one that follows a standard and solvable form, as we now integrate relatively simple expressions. Success in applying the substitution method rests on choosing the right substitution to transform the function effectively. This method greatly reduces the complexity of many integrals and is a staple in calculus for evaluating indefinite, definite, and iterated integrals alike.