Integration Techniques and Applications

Integration Techniques and Applications

Assessment

Interactive Video

Mathematics

9th - 12th Grade

Hard

Created by

Amelia Wright

FREE Resource

The video tutorial explains how to use integration by parts to find the antiderivative of a function. It begins by introducing the integration by parts formula and then guides the viewer through the process of selecting appropriate u and dv components. The tutorial demonstrates differentiating u and integrating dv, followed by applying the integration by parts formula to solve the integral. The video concludes with a simplified solution and a brief summary.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the formula for integration by parts?

The integral of u dv is equal to uv minus the integral of v du

The integral of u dv is equal to uv plus the integral of v du

The integral of u dv is equal to u plus v

The integral of u dv is equal to u minus v

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the integration by parts method, what is typically chosen as u?

The part of the integrand that makes differential u more complex

The part of the integrand that makes differential u simpler

The part of the integrand that is a constant

The part of the integrand that is a trigonometric function

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the derivative of 3x with respect to x?

1

3

0

x

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the integral of sine x with respect to x?

negative sine x

sine x

negative cosine x

cosine x

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the expression for v in the integration by parts formula when dv is sine x dx?

v equals negative sine x

v equals sine x

v equals cosine x

v equals negative cosine x

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the result of multiplying u and v in the integration by parts formula?

3x times cosine x

3x times negative cosine x

3x times sine x

3x times negative sine x

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What happens to the integral of v du when v is negative cosine x and du is 3 dx?

It becomes plus the integral of sine x dx

It becomes minus the integral of sine x dx

It becomes minus the integral of cosine x dx

It becomes plus the integral of cosine x dx

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