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Integration by Parts Concepts

Integration by Parts Concepts

Assessment

Interactive Video

Mathematics

11th Grade - University

Practice Problem

Hard

CCSS
8.EE.A.1, 6.EE.A.2C

Standards-aligned

Created by

Liam Anderson

FREE Resource

Standards-aligned

CCSS.8.EE.A.1
,
CCSS.6.EE.A.2C
The video tutorial explains how to evaluate a definite integral using the integration by parts method. It begins with an introduction to definite integrals and proceeds to demonstrate the integration by parts technique. The process is applied twice to simplify the integral, leading to the final antiderivative. The tutorial concludes with the evaluation of the definite integral and simplification of the result.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the purpose of using integration by parts in this problem?

To simplify the expression for easier differentiation

To eliminate the exponential term

To find the antiderivative of a complex integral

To convert the integral into a definite form

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the first application of integration by parts, what is chosen as 'u'?

The entire integral

Two x

x squared minus eight

e to the negative x

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the result of differentiating 'u' in the first integration by parts?

Two x dx

Negative e to the negative x

e to the negative x dx

x squared minus eight

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

During the second application of integration by parts, what is the new 'u'?

Negative e to the negative x

Two x

e to the negative x

x squared minus eight

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the integral of e to the negative x dx?

Negative e to the negative x

e to the x

x e to the negative x

Two e to the negative x

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the final form of the antiderivative before evaluating the definite integral?

The opposite of x squared minus eight times e to the negative x minus two x e to the negative x minus two e to the negative x

e to the negative x plus c

Negative 42 divided by e to the sixth

Two x e to the negative x plus c

Tags

CCSS.8.EE.A.1

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is the definite integral evaluated using the antiderivative?

By integrating the antiderivative again

By substituting the limits into the original integral

By setting the antiderivative equal to zero

By finding the difference between the antiderivative evaluated at the upper and lower limits

Tags

CCSS.6.EE.A.2C

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