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Understanding Double Integrals and Region of Integration

Understanding Double Integrals and Region of Integration

Assessment

Interactive Video

Mathematics

11th Grade - University

Practice Problem

Hard

CCSS
HSA.REI.D.12

Standards-aligned

Created by

Sophia Harris

FREE Resource

Standards-aligned

CCSS.HSA.REI.D.12
The video tutorial explains how to evaluate a double integral over a triangular region in the xy-plane. It begins by defining the function f(x, y) and the region's boundaries. The tutorial then sets up the double integral, initially integrating with respect to x and then y. Due to complexity, the order of integration is changed to integrate with respect to y first. The integral is evaluated using u-substitution, and the result is calculated as the volume under the surface above the xy-plane. The tutorial concludes with the exact and approximate values of the integral.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What are the boundaries of the triangular region in the xy-plane?

y = 1, x = 4, y = x

y = 0, x = 5, y = x

y = 1, x = 3, y = x

y = 0, x = 4, y = x

Tags

CCSS.HSA.REI.D.12

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What does the double integral over the region represent?

The slope of the surface

The perimeter of the region

The volume under the surface above the xy-plane

The area of the region

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

When integrating with respect to x first, what is the lower limit of integration?

x = 0

x = 4

x = y

x = 1

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why is it challenging to find the anti-derivative with respect to x in the original order of integration?

Because the integrand is a linear function

Because the integrand is a polynomial

Because the integrand involves a square root of x cubed plus one

Because the integrand is a constant

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the new order of integration after changing it?

Integrate with respect to x first, then y

Integrate with respect to y first, then x

Integrate with respect to z first, then y

Integrate with respect to y first, then z

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the lower limit of integration with respect to y in the new order?

y = 1

y = x

y = 4

y = 0

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What substitution is used to simplify the integral?

u = x to the fourth plus one

u = x squared plus one

u = x cubed plus one

u = x plus one

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