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 result is \(\frac{115}{6}\).

Step 1: Understanding the Problem

In this exercise, we are asked to evaluate the iterated integral \( \int_{-1}^{4} \int_{1}^{2} (x+y^2) \, dy \, dx \). This involves integrating with respect to \( y \) first, and then with respect to \( x \).

Step 1: Inner Integral with Respect to y

Integrate \(x + y^2\) with respect to \( y \) over the interval \([1, 2]\):\[\int_{1}^{2} (x + y^2) \, dy = \int_{1}^{2} x \, dy + \int_{1}^{2} y^2 \, dy.\]Integrating each part, we get:- The integral of \(x\) with respect to \(y\) is \(xy\), evaluated from 1 to 2: \([x(2) - x(1)] = x(2 - 1) = x\).- The integral of \(y^2\) with respect to \(y\) is \(\frac{y^3}{3}\), evaluated from 1 to 2: \(\left[\frac{8}{3} - \frac{1}{3}\right] = \frac{7}{3}\).Combining both results, the value of the inner integral is: \((x + \frac{7}{3})\).

Step 2: Evaluate the Outer Integral with Respect to x

Now, we integrate the result \((x + \frac{7}{3})\) with respect to \(x\) over the interval \([-1, 4]\):\[\int_{-1}^{4} \left(x + \frac{7}{3}\right) \, dx = \int_{-1}^{4} x \, dx + \int_{-1}^{4} \frac{7}{3} \, dx.\]Integrating each part, we get:- The integral of \(x\) with respect to \(x\) is \(\frac{x^2}{2}\), evaluated from -1 to 4: \(\left[\frac{16}{2} - \frac{1}{2}\right] = \frac{15}{2}\).- The integral of \(\frac{7}{3}\) with respect to \(x\) is \(\frac{7}{3}x\), evaluated from -1 to 4: \(\left[\frac{28}{3} + \frac{7}{3}\right] = \frac{35}{3}\).Combining both results: \(\frac{15}{2} + \frac{35}{3}\).

Step 3: Calculate the Final Result

Combine and simplify the results from Step 2:- Combine \(\frac{15}{2}\) and \(\frac{35}{3}\) by finding a common denominator, which is 6:- Convert \(\frac{15}{2}\) to \(\frac{45}{6}\) and \(\frac{35}{3}\) to \(\frac{70}{6}\).- Add the fractions: \(\frac{45}{6} + \frac{70}{6} = \frac{115}{6}\).Thus, the result of the original iterated integral is \(\frac{115}{6}\).

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics concerned with the accumulation of quantities and the areas under and between curves. It is one of the two main branches of calculus, with the other being differential calculus, which focuses on rates of change. At its core, integral calculus gives us tools to compute the area and cumulative quantities even when the shape does not have straight sides or constant rates.

When dealing with integrals, two terms are essential: the indefinite integral and the definite integral. An indefinite integral of a function represents a family of functions, while a definite integral computes the exact area under the curve of a function within a specific range.

In our exercise, the focus is on definite integrals, specifically iterated integrals. Iterated integrals involve integrating a function multiple times in a specified order. Here, we perform the integration with respect to one variable while treating others as constants, a technique commonly used in multivariable calculus.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions of several variables. This is an essential area of mathematics as it helps us to understand spaces and shapes with more than two dimensions. Functions often exist in multiple dimensions, such as a surface over a given region, which could represent physical quantities like mass, charge, or energy distribution.

In our exercise, we have a function, \(x + y^2\), involving two variables, \(x\) and \(y\). The iterated integral first considers integration with respect to \(y\), treating \(x\) as a constant, and then reverses the focus to integrate with respect to \(x\). This step-by-step integration across multiple dimensions characterizes multivariable calculus, enabling us to evaluate areas, volumes, and other such properties in higher-dimensional spaces.

Mastering these concepts can significantly advance your ability to solve complex real-world problems, involving anything from engineering systems to physics phenomena.

Definite Integrals

Definite integrals are used to find the signed area under a curve between two points on the x-axis. In mathematical notation, a definite integral of a function from \(a\) to \(b\) is denoted as \(\int_{a}^{b} f(x) \, dx\). It provides a numerical value representing this total accumulation or reduction within the specified limits.

In iterated integrals, definite integrals are taken over multiple dimensions, progressively collapsing a multidimensional space into a single value through repeated integration. As in our problem, first, we compute the integral from 1 to 2 with respect to \(y\), resulting in a function of \(x\).

Next, we integrate the resulting expression over the interval from \(-1\) to \(4\) concerning \(x\). This two-step process allows us to evaluate the total value across a region defined horizontally and vertically by these boundaries, refining our result into a single number, \(\frac{115}{6}\).

• This approach involves the Fubini's Theorem, which assists in calculating multiple integrals by iterating single-variable integrals.

Understanding these processes enables you to handle further intricate integrals in multivariable calculus.