/*! 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 17 Evaluate the double integral. ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 17

Evaluate the double integral. $$ \int_{1}^{2} \int_{0}^{4}\left(3 x^{2}-2 y^{2}+1\right) d x d y $$

Short Answer

Expert verified
The value of the double integral is 56

Step by step solution

01

Integrate with respect to x

Integral of \(3x^2\) with respect to x is \(x^3\), integral of \(2y^2\) with respect to x is \(2y^2x\) (note that y is treated as a constant), integral of 1 with respect to x is x. Thus the integral becomes: \[\int_{1}^{2} [x^3 - 2y^2x + x]_{x=0}^{x=4} dy\] Evaluate the values of x at 0 and 4, subtracting the lower limit from the upper limit for each.
02

Evaluate x Integral

After substituting x=4 and x=0 into [x^3 - 2y^2x + x], we get: \[\int_{1}^{2} (64 - 8y^2 + 4) dy\] This results in \[\int_{1}^{2} (68 - 8y^2) dy\] Now we are ready to proceed to integrate with respect to y.
03

Integrate with respect to y

The integral of 68 with respect to y is 68y, and the integral of -8y^2 with respect to y is \(- \frac{8y^3}{3}\). Thus the integral becomes: \[[68y - \frac{8y^3}{3}]_{y=1}^{y=2}\] . Evaluate the values of y at 1 and 2, subtracting the lower limit from the upper limit for each.
04

Evaluate Final Result

After substituting y=2 and y=1 into [68y - \frac{8y^3}{3}], we get \(136 - \frac{64}{3}\) - \(68 - \frac{8}{3})\). Simplifying gives the final result.

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!

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

Use the regression capabilities of a graphing utility or a spreadsheet to find the least squares regression line for the given points. $$ (-3,4),(-1,2),(1,1),(3,0) $$

Use a double integral to find the area of the region bounded by the graphs of the equations. $$ y=x, y=2 x, x=2 $$

Use a double integral to find the area of the region bounded by the graphs of the equations. $$ y=9-x^{2}, y=0 $$

Use a double integral to find the area of the region bounded by the graphs of the equations. $$ y=x^{3 / 2}, y=x $$

Plot the points and determine whether the data have positive, negative, or no linear correlation (see figures below). Then use a graphing utility to find the value of \(r\) and confirm your result. The number \(r\) is called the correlation coefficient. It is a measure of how well the model fits the data. Correlation coefficients vary between \(-1\) and 1, and the closer \(|r|\) is to 1, the better the model. $$ (0.5,9),(1,8.5),(1.5,7),(2,5.5),(2.5,5),(3,3.5) $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Statistics

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.