/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Evaluate the iterated integral \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 1

Evaluate the iterated integral \(\int_{0}^{4} \int_{0}^{4} 12 x^{2} y^{3} d x d y\)

Short Answer

Expert verified
We evaluate the given iterated integral \(\int_{0}^{4} \int_{0}^{4} 12 x^{2} y^{3} d x d y\) sequentially by first integrating with respect to \(x\), then with respect to \(y\). The inner integral evaluates to \(64y^3\), and the outer integral evaluates to \(16,384\). So, the value of the iterated integral is \(16,384\).

Step by step solution

01

Identify the order of integration

The given integral is in the form: \[\int_{0}^{4} \int_{0}^{4} 12 x^{2} y^{3} d x d y\] First, we will integrate the inner integral concerning \(x\), and then proceed to integrate the outer integral concerning \(y\).
02

Evaluate the inner integral with respect to \(x\)

We begin by integrating the given function with respect to \(x\): \[\int_{0}^{4} 12 x^2 y^3 dx\] Applying the power rule of integration, we get, \[\frac{12}{3} x^3 y^3 \Big|_0^4\] Now evaluate this by substituting the limits of integration: \[\frac{12}{3} (4^3 y^3 - 0^3 y^3)\] This reduces to: \[64 y^3\]
03

Evaluate the outer integral with respect to \(y\)

Now, integrate the result obtained in the previous step with respect to \(y\): \[\int_{0}^{4} 64 y^3 dy\] Applying the power rule of integration again, we obtain: \[\frac{64}{4} y^4 \Big|_0^4\] Substituting the limits of integration: \[\frac{64}{4} (4^4 - 0^4)\] This reduces to: \[64(256)\]
04

Calculate the final result

Finally, we compute the result: \[64(256) = 16384\] So, the value of the iterated integral is \(16,384\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration
Integration is a fundamental concept in calculus that allows us to find areas under curves, among other applications. It can be thought of as an operation that reverses differentiation. In the context of our exercise, integration is used to evaluate the iterated integral, which involves calculating a double integral.
  • In our example, the integral is performed in two stages: first with respect to one variable, followed by with respect to another variable. This process is known as iterated integration.
  • The primary function is expressed as an algebraic function within the integral, specifically as a polynomial, which is generally straightforward to integrate using established rules.
Understanding the process and mechanics of integration is essential in tackling problems involving areas, volumes, and other applications in physics and engineering.
Power Rule of Integration
The power rule of integration is a handy tool for finding the integral of a polynomial expression. If you are familiar with the power rule of differentiation, this rule is quite similar but in reverse.
  • For any term of the form \( ax^n \), the integral is given by \( \frac{a}{n+1}x^{n+1} + C \), where \( C \) is the constant of integration.
  • In the iterated integral problem, we applied the power rule multiple times to handle the polynomial components \( 12x^2y^3 \) and then \( 64y^3 \). With each application, the exponent of the variable is increased by one, and the new expression is divided by this new exponent.
  • This rule simplifies many integration problems, making the integration of simple polynomials straightforward and systematic.
Multiple Integration
Multiple integration extends the idea of a regular integral to functions of multiple variables. It involves the integration of functions over regions within a plane or space, often denoted by double or triple integrals.
  • In our exercise, we focused on a double integral, which required integrating with respect to two variables: \( x \) and \( y \).
  • The iterated integral \( \int_{0}^{4} \int_{0}^{4} 12 x^2 y^3 \ dx \ dy \) demonstrates how to handle two integrals separately, but in sequence. This sequential approach helps in simplifying complex multivariable integrations.
  • Multiple integrals are crucial in computing areas, volumes, center of mass, and other applications in higher dimensions, providing a powerful technique to analyze multi-dimensional spaces.
Calculus
Calculus is a branch of mathematics that studies continuous change. It comprises two main areas: differentiation and integration.
  • Differentiation involves finding the rate of change and slopes of curves, while integration involves finding the total accumulation or area under curves.
  • Within our exercise, calculus is used to solve an iterated integral, showcasing the integration aspect to find a specific area bound by the limits \( 0 \) and \( 4 \) for both variables, \( x \) and \( y \).
  • The fundamental theorem of calculus connects these two ideas, stating that differentiation and integration are inverse processes. This duality forms a bedrock for evaluating definite integrals, like the one in our problem.
By embracing the fundamentals of calculus, students can unlock a deeper understanding of motion, growth, decay, and many other phenomenons described by mathematical models.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the average value of \(f(x, y)=4 x^{6} y^{3}\) over the rectangle \(\mathrm{R}\) with vertices (-3,0),(-3,6),(3,0),(3,6) . Average value \(=\) ______________________.

Consider the iterated integral \(I=\int_{x=0}^{x=4} \int_{y=\sqrt{x}}^{y=2} \cos \left(y^{3}\right) d y d x\). a. Sketch the region of integration. b. Write an equivalent iterated integral with the order of integration reversed. c. Choose one of the two orders of integration and evaluate the iterated integral you chose by hand. Explain the reasoning behind your choice. d. Determine the exact average value of \(\cos \left(y^{3}\right)\) over the region \(D\) that is determined by the iterated integral \(I\).

Find the centroid \((\bar{x}, \bar{y})\) of the triangle with vertices at \((0,0),(5,0),\) and (0,2) \(\bar{x}=\) \(\bar{y}=\)

Evaluate each of the following double or iterated integrals exactly. a. \(\int_{1}^{3}\left(\int_{2}^{5} x y d y\right) d x\) b. \(\int_{0}^{\pi / 4}\left(\int_{0}^{\pi / 3} \sin (x) \cos (y) d x\right) d y\) c. \(\int_{0}^{1}\left(\int_{0}^{1} e^{-2 x-3 y} d y\right) d x\) d. \(\iint_{R} \sqrt{2 x+5 y} d A,\) where \(R=[0,2] \times[0,3]\).

A lamp has two bulbs, each of a type with an average lifetime of 10 hours. The probability density function for the lifetime of a bulb is \(f(t)=\) \(\frac{1}{10} e^{-t / 10}, t \geq 0\) What is the probability that both of the bulbs will fail within 3 hours?
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.