Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each of the iterated integrals. \(\int_{-1}^{4} \int_{1}^{2}\left(x+y^{2}\right) d y d x\)

Step-by-Step Solution

Answer

The value of the iterated integral is \( \frac{115}{6} \).

Step 1: Set up the Outer Integral

The given integral is iterated, which means we first integrate with respect to the inner variable, and then with respect to the outer variable. Identify the outer integral: \( \int_{-1}^{4} [f(x)] \, dx \). The limits for \( x \) range from -1 to 4.

Step 2: Integrate with Respect to y

Focus on the inner integral: \( \int_{1}^{2} (x+y^2) \, dy \). When integrating \( x \) with respect to \( y \), treat \( x \) as a constant. Thus:\[ \int (x+y^2) \, dy = xy + \frac{y^3}{3} + C \]Now evaluate this from \( y=1 \) to \( y=2 \).

Step 3: Evaluate the Inner Definite Integral

Substitute the limits for \( y \):- When \( y = 2 \), the expression becomes \( x(2) + \frac{2^3}{3} = 2x + \frac{8}{3} \).- When \( y = 1 \), the expression becomes \( x(1) + \frac{1^3}{3} = x + \frac{1}{3} \).Subtract the two results: \[(2x + \frac{8}{3}) - (x + \frac{1}{3}) = x + \frac{7}{3}\]Now the expression inside the outer integral becomes \( \int_{-1}^{4} (x + \frac{7}{3}) \, dx \).

Step 4: Integrate with Respect to x

Now integrate the expression \( x + \frac{7}{3} \) with respect to \( x \):\[\int (x + \frac{7}{3}) \, dx = \frac{x^2}{2} + \frac{7}{3}x + C\]Evaluate this from \( x = -1 \) to \( x = 4 \).

Step 5: Evaluate the Outer Definite Integral

Substitute the limits for \( x \):- When \( x = 4 \), the result is \( \frac{4^2}{2} + \frac{7}{3} \cdot 4 = 8 + \frac{28}{3} \).- When \( x = -1 \), the result is \( \frac{(-1)^2}{2} + \frac{7}{3} \cdot (-1) = \frac{1}{2} - \frac{7}{3} \).Calculate:- \( 8 + \frac{28}{3} = \frac{24}{3} + \frac{28}{3} = \frac{52}{3} \).- \( \frac{1}{2} - \frac{7}{3} = \frac{3}{6} - \frac{14}{6} = -\frac{11}{6} \).Now find \( \frac{52}{3} + \frac{11}{6} \), which requires a common denominator:The result is \( \frac{104}{6} + \frac{11}{6} = \frac{115}{6} \).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, specifically when it comes to evaluating the area under a curve between two points. The concept involves two key components: a function to integrate and specific limits of integration.

• The integral sign \( \int \) tells us to perform integration.
• The limits, for example, \( a \) to \( b \), define the interval over which we integrate.
• A key output of a definite integral is a numerical value representing the total accumulation (like area under a curve) over a given range.

When solving an iterated (or multiple) integral like \( \int_{-1}^{4} \int_{1}^{2} (x+y^2) \, dy \, dx \), it's essential to understand this layered integration process. In our exercise, we first calculate the integral with respect to \( y \) while keeping \( x \) constant, giving us a new function. Then, we evaluate the resulting function over the \( x \) limits to get our final solution. This multi-stage approach allows us to simplify and solve complex problems involving more than one variable.

Integration Techniques

When tackling iterated integrals, it's crucial to employ various integration techniques to simplify the problem. The technique used in the exercise is the separation of integrations by recognizing one variable as constant at each stage.

• First, integrate with respect to \( y \), the inner variable, treating \( x \) as a constant. For instance, integrating \( (x + y^2) \) with respect to \( y \) results in \( xy + \frac{y^3}{3} \).
• Evaluate this expression between the bounds for \( y \), which transforms our integrand into a simpler form for the next step.
• Proceed to integrate the result with respect to \( x \), the outer variable, using standard techniques like polynomial integration.

Using each technique effectively builds a comprehensive solution. Keeping these steps clear and separate prevents confusion, allowing for systematic and repeatable solutions.

Multivariable Calculus

Multivariable calculus extends basic calculus concepts to functions of more than one variable. It addresses questions about finding areas, volumes, and other multi-dimensional constructs, such as seen in iterated integrals. In this context, multivariable calculus allows a deeper exploration of the relationships between variables and their effects on integrations.

• For instance, in an expression with \( x \) and \( y \), we explore their combined roles in determining the volume under a surface above a region in the \( xy \)-plane.
• Iterated integrals, as in our problem, illustrate a practical application where two variables interact within defined bounds.
• Each step compresses part of the complexity into manageable calculations by separating interactions of variables, allowing a focus on the process of integration for each variable distinctly.

Understanding iterated integrals is vital in multivariable calculus, as it builds the foundation for solving real-world problems, such as calculating mass, center of mass, or electrical charge distributions across a plane.