Integration by Substitution Concepts

Integration by Substitution Concepts

Assessment

Interactive Video

Mathematics

10th - 12th Grade

Practice Problem

Hard

Created by

Mia Campbell

FREE Resource

The video tutorial explains how to evaluate an indefinite integral using integration by substitution. It begins by identifying the need for substitution and choosing an appropriate substitution variable. The process involves rewriting the integral in terms of the new variable, solving it, and then converting back to the original variable. The tutorial concludes with a complete solution and a preview of another example in the next video.

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9 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the main goal of using integration by substitution?

To solve differential equations

To differentiate the function

To simplify the integral by changing variables

To find the limit of the function

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

When using integration by substitution, what is the first step?

Differentiate the function

Choose a substitution variable 'u'

Integrate directly

Find the derivative of 'u'

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the substitution method, what does 'du' represent?

The constant of integration

The original function

The integral of 'u'

The derivative of 'u' with respect to 'x'

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How do you handle the differential 'dx' in the substitution process?

Multiply it by 'u'

Express 'dx' in terms of 'du'

Ignore it

Replace it with 'du' directly

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the antiderivative of 1/u with respect to 'u'?

1/u + C

u^2/2

e^u + C

ln|u| + C

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

After finding the antiderivative in terms of 'u', what is the next step?

Find the limit

Convert back to the original variable 'x'

Differentiate the result

Solve for 'u'

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the final form of the antiderivative in terms of 'x' for the given example?

ln|5x + 1| + C

5/2 ln|x - 1| + C

2/5 ln|5x - 1| + C

2 ln|x - 5| + C

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