Integration by Parts

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 7.1 – Integration by Parts

Expected Educational Results

• Objective 03–01: I can use integration by parts to evaluate an indefinite integral using FTC-I.

• Objective 03–02: I can use integration by parts to evaluate a definite integral using FTC-II.

• Objective 03–03: I can use substitution and integration by parts to evaluate an indefinite integral using FTC-I.

• Objective 03–04: I can use substitution and integration by parts to evaluate a definite integral using FTC-II.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Integration by Parts

Definition: Integration by Parts

If $U=f(x)$$U=f(x)$ is a differentiable function and $dV=g^{\prime }(x)$$dV=g^{\,\prime}(x)$ is a integrable function, then $dU=f^{\prime }(x)$$dU=f^{\,\prime}(x)$ and $V=g(x)$$V=g(x)$. Thus, $\int UdV=UV-\int VdU$$\int{U\,dV}=UV-\int{V\,dU}$

Procedure: Traditional Method for Integration by Parts

When using Integration by Parts, the typical first step is:

• Let $U=\text{polynomial function}$$U=\text{polynomial function}$

• $dV=\text{everything else}$$dV=\text{everything else}$

NOTE: Sometimes the above first step does not work. We will discuss this case later in this Module.

Example 01: Evaluate: ${\displaystyle \int x{e}^{x}dx}$$\displaystyle\int{xe^x\,dx}$

Solution 01: The polynomial factor is $x$$x$. So, let $U=x$$\color{red}U=x$, then let $dV=e^{x}dx$$\color{blue}dV=e^x\,dx$.

We need $dU$$dU$ and $V$$V$:

$dU=dx$$\color{orange}dU = dx$ $V=e^x$$\color{purple}V=e^x$

NOTE: There is NO $+C$$+C$ included with $V$$V$.

Now substitute $U$$U$, $dU$$dU$, $V$$V$, and $dV$$dV$ into Definition shown above:

$\int x{e}^{x}dx=\int UdV$$\int{xe^x\,dx} = \int{UdV}$

$\Rightarrow UV-\int VdU$$\Rightarrow UV - \int{VdU}$

$\Rightarrow x{e}^x-\int e^{x}dx$$\Rightarrow {\color{red}x} {\color{purple}e^x} - \int{{\color{purple}e^x}\,{\color{orange}dx}}$

NOTE: Since there is an integral remaining in the line above, there is no $+C$$+C$.

The above line still has an integral. So we need to evaluate the integral term, $\int e^{x}dx$$\int{{\color{purple}e^x}\,{\color{orange}dx}}$.

$\Rightarrow x{e}^x-e^x+C$$\Rightarrow x e^x - e^x + C$ ∎

NOTE: Since there is no integral remaining in the line above, you must include $+C$$+C$.

Example 02: Evaluate: ${\displaystyle \int x^2{e}^{x}dx}$$\displaystyle\int{x^2 e^x\,dx}$​​

Solution 02a: The polynomial factor is $x^2$$x^2$​​. So, let $U=x^2$$U=x^2$​​, then let $dV=e^{x}dx$$dV=e^x\,dx$​​.

We need $dU$$dU$ and $V$$V$:

$dU=2xdx$$dU = 2x\,dx$​ $V=e^x$$V=e^x$​

NOTE: There is NO $+C$$+C$ included with $V$$V$.

Now substitute $U$$U$, $dU$$dU$, $V$$V$, and $dV$$dV$ into Definition shown above:

$\int x^2{e}^{x}dx=\int UdV$$\int{x^2 e^x\,dx} = \int{UdV}$​

$\Rightarrow UV-\int VdU$$\Rightarrow UV - \int{VdU}$

$\Rightarrow x^2{e}^x-\int 2x{e}^{x}dx$$\Rightarrow x^2 e^x - {\color{red}\int{2x e^x\,dx}}$​

NOTE: We need to integrate (the red integral term) by parts again!

The polynomial factor is $2x$$2x$​​. So, let $U=2x$$U=2x$​​, then let $dV=e^{x}dx$$dV=e^x\,dx$​​.

We need $dU$$dU$ and $V$$V$:

$dU=2dx$$dU = 2\,dx$​​ $V=e^x$$V=e^x$​​

NOTE: There is NO $+C$$+C$ included with $V$$V$.

Now substitute $U$$U$, $dU$$dU$, $V$$V$, and $dV$$dV$ into Definition shown above:

$\int x^2{e}^{x}dx=x^2{e}^x-\int UdV$$\int{x^2 e^x\,dx} = x^2 e^x - {\color{red}\int{UdV}}$​​​

$\Rightarrow x^2{e}^x-\left(UV-\int VdU\right)$$\Rightarrow x^2 e^x - \left({\color{red}UV - \int{VdU}}\right)$​​​

$\Rightarrow x^2{e}^x-\left(2x{e}^x-\int 2e^{x}dx\right)$$\Rightarrow x^2 e^x - \left({\color{red}2x e^x - \int{2 e^x\,dx}}\right)$​

Evaluate the last integral term (include $+C$$+C$ here) and distribute the negative sign:

$\Rightarrow x^2{e}^x-2x{e}^x+2e^x+C$$\Rightarrow x^2 e^x - 2x e^x + 2e^x + C$ ∎

Solution 02b:

Procedure: The Tanzalin Method for Integration by Parts

For a discussion of this method: www.intmath.com/blog/mathematics/tanzalin-method-for-easier-integrationby-parts-4339.

Multiply along the diagonals and use alternating signs along diagonals.

${\displaystyle \int x^2{e}^{x}dx=x^2{e}^x-2x{e}^x+2e^x+C}$$\displaystyle\int{x^2 e^x dx}= x^2 e^x - 2x e^x + 2e^x + C$ ∎

Investigation 04

1. Compare the result of integrating ${\displaystyle \int x^2{e}^{x}dx}$$\displaystyle\int{x^2 e^x dx}$​​ using the Definition of Integration by Parts versus using the Tanzalin Method.

2. Explain the use of the $+$$+$ and $-$$-$ signs along the diagonals in the Tanzalin Method.

Using Integration by Parts for Indefinite Integrals

Investigation 05

Use either procedure for Integration by Parts to integrate the following:

1. ${\displaystyle \int x{e}^{x}dx}$$\displaystyle\int{xe^x\,dx}$

2. ${\displaystyle \int x\sin (x)dx}$$\displaystyle\int{x\sin{(x)}\,dx}$

3. ${\displaystyle \int x^2\cos \left(2x\right)dx}$$\displaystyle\int{x^2\cos{\left(2x\right)}\,dx}$

4. ${\displaystyle \int x^3{e}^{6x}dx}$$\displaystyle\int{x^3e^{6x}\,dx}$

5. ${\displaystyle \int x^2{e}^{x}dx}$$\displaystyle\int{x^2e^x\,dx}$

Use Technology to Verify Antiderivatives

Mathematica

NOTE: Mathematica is used here to verify antiderivatives. You must show that you understand the Calculus concept of the antiderivative.

1
(* Example from Example 02: Integrate x^2e^x *)
2
Integrate[x^2*E^x, x]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: https://www.jjw3.com/Common_Mathematica_Code.html.

2. The Mathematica syntax for ${\sin }^{-1}(x)$$\sin^{-1}{(x)}$ is $\text{ArcSin[x]}$$\text{ArcSin[x]}$, $e^x$$e^x$ is $\text{E}\wedge x$$\text{E}\!\wedge\!x$, ${\sec }^2(x)$$\sec^2{(x)}$ is $\text{Sec[x]}\wedge 2$$\text{Sec[x]}\!\wedge\!2$​​, etc.

3. Remember to use parens $($$\left(\right.$ and $)$$\left.\right)$ to group multiple terms in the numerator or denominator of a rational expression.

4. Integrate[ ] for indefinite integrals has two arguments:

   1. The integrand, i.e., function of the definite integral;

   2. $x$$x$​, where $x$$x$​ is the variable used in the integrand.

5. Remember, correct Mathematica code will be all black except for variables.

6. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

7. Mathematica does not show $+C$$+C$ for antiderivatives. You will need to remember to include $+C$$+C$ in your answers.

8. Mathematica may return a result in a different, but equivalent, form than your answer. In these cases, use FTC-I to check your work.

Python

NOTE: Python is used here to verify definite integrals. On all Assessments, you must show that you understand the Calculus I concept of using $u$$u$-Substitution to evaluate an indefinite integral.

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Python syntax is located at: https://www.jjw3.com/Common_Python_Code.html.

2. All code is included above with comment codes that explain the executable code.

3. integrate( ) for indefinite integrals has two arguments:

   1. The integrand, i.e., function of the definite integral;

   2. $x$$x$​, where $x$$x$​​ is the variable used in the integrand.

4. You only need to change the function in the integrate( ) command.

5. The Python syntax for $e^x$$e^x$​​ is $\text{exp(x)}$$\text{exp(x)}$​​; ${\cos }^2(x)$$\cos^2{(x)}$​​ is $\text{cos(x)**2}$$\text{cos(x)**2}$​​; $x^4$$x^4$​​ is $\text{x**4}$$\text{x**4}$​​, $\ln (x)$$\ln{(x)}$​​ is $\text{log(x)}$$\text{log(x)}$​​, ${\tan }^{-1}(x)$$\tan^{-1}{(x)}$​​ is $\text{atan(x)}$$\text{atan(x)}$​​, etc.

6. Python requires $\ast$$*$ for explicit multiplication; $\ast \ast$$**$​ for exponents.

7. Remember to use parens $($$\left(\right.$ and $)$$\left.\right)$​ to group multiple terms in the numerator or denominator of a rational expression.

8. Python does not return the antiderivative with $+C$$+C$. You need to use $+C$$+C$ for all steps that are antiderivatives.

9. To execute code, press the “View the result” button:

Homework

At this time, you should be able to complete the following assignments:

• Section 7.1: # 1, 3, 5, 11, 35.

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Created: Thursday, 12 August 2021 - 15:26 (EDT) Last Modified: Tuesday, 16 August 2022 - 15:34 (EDT)