<meta http-equiv="refresh" content="1; url=/nojavascript/"> Integration by Parts | CK-12 Foundation
Dismiss
Skip Navigation
You are reading an older version of this FlexBook® textbook: CK-12 Texas Instruments Calculus Student Edition Go to the latest version.

This activity is intended to supplement Calculus, Chapter 7, Lesson 2.

In this activity, you will explore:

  • product rule of differentiation
  • integration by parts

Use this document to record your answers.

Exercises

1. State the product rule for a function of the form u(x)*v(x).

2. Apply the product rule to the function \sin(x)* \ln(x).

3. Do you agree or disagree with the following statement? Explain.

\int\limits \frac{d}{dx} (f(x)) dx = \frac{d}{dx} \left (\int\limits f(x)dx \right ) = f(x)

4. What is the integral of the left side of the product rule?

\int\limits \left (\frac{d}{dx} (u(x) \cdot v(x) \right )dx =

5. What is the integral of the right side?

\int\limits \left (u (x) \cdot \frac{dv}{dx} + v(x) \cdot \frac{du}{dx} \right )dx =

6. Explain the relationship between the areas shown on the graph and the following equation:

\int\limits_{v_1}^{v_2} u \cdot dv = u \cdot v - \int\limits_{u_1}^{u_2} v \cdot du

7. Use the method of integration by parts to compute the integral of \ln (x).

Remember the formula for Integration by parts is \int\limits u \cdot dv = u \cdot v - \int\limits v \cdot du

& \int\limits \ln(x) \cdot 1\ dx \rightarrow u = \ln(x) \ \text{and} \ dv = 1\ dx \\& \qquad \qquad \qquad \quad \ du = \qquad \qquad \quad v = \\& \text{Result} =

Check by integration directly. (Home > F3:Calc > 2:Integrate) or (Home > 2^{nd} 7 )

Consider the function f(x) = \sin(\ln(x)).

u & = \sin(\ln(x)) \rightarrow du = \frac{\cos(\ln(x))}{x} dx \\dv & = dx \rightarrow v = x (+C)

\int\limits \sin(\ln(x)) \cdot 1\ dx & = x \cdot \sin(\ln(x)) - \int\limits  x \cdot \frac{\cos(\ln(x))}{x} dx (+C) \\& = x \cdot \sin(\ln(x)) - \int\limits \cos (\ln(x))dx (+C)

8. Find \int\limits \cos (\ln(x))dx.

u & = && du = && dv = && v = \\\int\limits \cos (\ln(x))dx  & =

9. Substitute the result for \cos(\ln(x)) into the result for \sin(\ln(x)).

u & = && du = && dv = && v = \\\int\limits \sin (\ln(x))dx  & =

10. Use integration by parts to solve the following. If you need to use integration by parts more than once, do so. Check your result.

a. \int\limits \tan^{-1}(x)\ dx

b. \int\limits x^2 \cdot e^x \ dx

c. \int\limits  x \cdot \tan^{-1}(x) \ dx

d. \int\limits  x \cdot \cos (2x+1)\ dx

11. (Extension 1) Does it matter in which order u(x) and v(x) are selected for the method of integration by parts?

12. (Extension 2) Is there likely to be an integration rule based upon the quotient rule just as Integration by Parts was based upon the product rule?

Image Attributions

Description

Categories:

Grades:

Date Created:

Feb 23, 2012

Last Modified:

Aug 19, 2014
Files can only be attached to the latest version of None

Reviews

Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
TI.MAT.ENG.SE.1.Calculus.8.2
ShareThis Copy and Paste

Original text