Session 06-05 - Integration by Parts & Synthesis

Section 06: Integral Calculus

Author

Dr. Nikolai Heinrichs & Dr. Tobias Vlcek

Entry Quiz - 10 Minutes

Quick Review from Session 06-04

Test your understanding of Area Between Curves & Surplus

Find the area between \(f(x) = x^2\) and \(g(x) = 2x\) from \(x = 0\) to \(x = 2\).
If demand is \(D(q) = 80 - 2q\) and supply is \(S(q) = 20 + q\), find equilibrium \((q^*, p^*)\).
Set up (but don’t evaluate) the consumer surplus integral for the market in problem 2.
Find \(\int (3x^2 - 4x + 5) \, dx\)

Homework Discussion - 15 Minutes

Your questions from Session 06-04

Focus on area between curves and economic surplus

Finding intersection points
Consumer and producer surplus calculations
Price floor/ceiling effects on surplus
Deadweight loss interpretation

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Today we learn a powerful technique for integrating products of functions and synthesize our integration skills!

Learning Objectives

What You’ll Master Today

Understand integration by parts as the reverse of the product rule
Apply the LIATE rule to choose which function to differentiate
Solve integrals of the form \(\int x \cdot e^x \, dx\)
Handle repeated application for integrals like \(\int x^2 \cdot e^x \, dx\)
Compute average value of a function over an interval
Apply integration to revenue and cost accumulation
Synthesize techniques from differential and integral calculus

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Integration by parts is essential for the Feststellungsprufung!

Part A: The Product Rule in Reverse

Motivation: Why Do We Need This?

Consider this integral:

\[\int x \cdot e^x \, dx\]

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Question: Can we use the power rule?

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No!

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Question: Can we use substitution?

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Try \(u = x\) or \(u = e^x\)… Neither works cleanly!

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Some products of functions require a special technique.

The Product Rule Revisited

Recall the product rule for derivatives:

\[\frac{d}{dx}[u \cdot v] = u' \cdot v + u \cdot v'\]

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Rearranging:

\[u \cdot v' = \frac{d}{dx}[u \cdot v] - u' \cdot v\]

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Integrating both sides:

\[\int u \cdot v' \, dx = u \cdot v - \int u' \cdot v \, dx\]

The Integration by Parts Formula

\[\int \underbrace{u(x)}_{\text{choose}} \cdot \underbrace{v'(x) \, dx}_{= \, dv} = u(x) \cdot v(x) - \int \underbrace{v(x)}_{\text{from integrating } dv} \cdot \underbrace{u'(x) \, dx}_{= \, du}\]

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This is the same as:

\[\int u \, dv = uv - \int v \, du\]

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Strategy: Choose \(u\) and \(dv\) so that \(\int v \, du\) is easier than \(\int u \, dv\)!

What Do \(du\) and \(dv\) Mean?

You already know this notation — from every integral you’ve written!

When you write \(\int x^2 \, dx\), the \(dx\) tells you what variable you integrate with respect to
Now, \(du\) works the same way — it just says “with respect to \(u\)”
The connection between them: if \(u\) is a function of \(x\), we can convert between \(du\) and \(dx\): \(du = u'(x) \, dx\)
In words: \(du\) is the derivative of \(u\) with respect to \(x\), multiplied by \(dx\)
The \(dx\) is needed for the remaining integral, don’t worry about it

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Question: Do you get the basic idea?

Part B: The LIATE Rule

Choosing \(u\) - The LIATE Rule

How do we choose which function to call \(u\)?

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Letter	Function Type	Example
L	Logarithmic	\(\ln(x)\), \(\log(x)\)
I	Inverse trig	\(\arctan(x)\), \(\arcsin(x)\)
A	Algebraic	\(x\), \(x^2\), polynomials
T	Trigonometric	\(\sin(x)\), \(\cos(x)\)
E	Exponential	\(e^x\), \(2^x\)

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Choose \(u\) in this order of priority!

Why LIATE Works

The idea: Choose \(u\) to be the function that becomes simpler when differentiated.

Logarithms (\(\ln x \to \frac{1}{x}\)): Derivatives are simpler
Algebraic (\(x^n \to nx^{n-1}\)): Powers decrease
Exponentials (\(e^x \to e^x\)): Don’t simplify - put in \(dv\)!

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The LIATE rule is a guideline, not a strict rule. Sometimes you need to experiment!

Part C: Basic Examples

Example 1: \(\int x \cdot e^x \, dx\)

Step 1: Identify \(u\) and \(dv\) using LIATE: choose \(u = x\) (A comes before E)

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Step 2: Set up the parts

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We choose:	We compute:
\(u = x\)	\(du = 1 \cdot dx = dx\) (differentiate \(u\), attach \(dx\))
\(dv = e^x \, dx\)	\(v = \int e^x \, dx = e^x\) (integrate \(dv\) to get \(v\))

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Step 3: Apply the formula \(\int u \, dv = uv - \int v \, du\)

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\[\int \underbrace{x}_{u} \cdot \underbrace{e^x \, dx}_{dv} = \underbrace{x}_{u} \cdot \underbrace{e^x}_{v} - \int \underbrace{e^x}_{v} \, \underbrace{dx}_{du}\]

Example 1: Solution

\[\int x \cdot e^x \, dx = x \cdot e^x - \int e^x \, dx\]

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\[= x \cdot e^x - e^x + C\]

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\[= e^x(x - 1) + C\]

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Verification: Differentiate to check: \[\frac{d}{dx}[e^x(x-1)] = e^x(x-1) + e^x \cdot 1 = e^x \cdot x - e^x + e^x = x \cdot e^x \quad \checkmark\]

Example 2: \(\int (x+1) \cdot e^x \, dx\)

This is a common exam problem!

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Step 1: Choose \(u = x + 1\) (algebraic), \(dv = e^x \, dx\)

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We choose:	We compute:
\(u = x + 1\)	\(du = 1 \cdot dx = dx\) (differentiate \(u\), attach \(dx\))
\(dv = e^x \, dx\)	\(v = \int e^x \, dx = e^x\) (integrate \(dv\) to get \(v\))

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Step 2: Apply the formula

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\[\int (x+1) \cdot e^x \, dx = (x+1) \cdot e^x - \int e^x \, dx\]

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\[= (x+1) \cdot e^x - e^x + C = e^x(x + 1 - 1) + C = x \cdot e^x + C\]

Example 3: \(\int x \cdot \ln(x) \, dx\)

Step 1: Identify using LIATE: choose \(u = \ln(x)\) (L comes first!)

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Step 2: Set up the parts

We choose:	We compute:
\(u = \ln(x)\)	\(du = \frac{1}{x} \cdot dx\) (differentiate \(u\), attach \(dx\))
\(dv = x \, dx\)	\(v = \int x \, dx = \frac{x^2}{2}\) (integrate \(dv\) to get \(v\))

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Step 3: Apply the formula

\[\int x \cdot \ln(x) \, dx = \ln(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx\]

Example 3: Solution

\[\int x \cdot \ln(x) \, dx = \frac{x^2 \ln(x)}{2} - \int \frac{x}{2} \, dx\]

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\[= \frac{x^2 \ln(x)}{2} - \frac{x^2}{4} + C\]

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\[= \frac{x^2}{4}\left(2\ln(x) - 1\right) + C\]

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Notice how choosing \(u = \ln(x)\) simplified the integral because \(\frac{1}{x} \cdot x^2 = x\) (simpler than before).

Break - 10 Minutes

Part D: Repeated Integration by Parts

When One Application Isn’t Enough

Consider: \(\int x^2 \cdot e^x \, dx\)

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First application: \(u = x^2\), \(dv = e^x \, dx\)

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We choose:	We compute:
\(u = x^2\)	\(du = 2x \cdot dx\) (differentiate \(u\), attach \(dx\))
\(dv = e^x \, dx\)	\(v = \int e^x \, dx = e^x\) (integrate \(dv\) to get \(v\))

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\[\int x^2 \cdot e^x \, dx = x^2 \cdot e^x - \int 2x \cdot e^x \, dx\]

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Problem: We still have \(\int x \cdot e^x \, dx\) - another integration by parts problem!

Repeated Application

Second application: For \(\int x \cdot e^x \, dx\), let \(u = x\), \(dv = e^x \, dx\)

\[\int x \cdot e^x \, dx = x \cdot e^x - e^x\]

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Substituting back:

\[\int x^2 \cdot e^x \, dx = x^2 \cdot e^x - 2(x \cdot e^x - e^x) + C\]

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\[= x^2 \cdot e^x - 2x \cdot e^x + 2e^x + C= e^x(x^2 - 2x + 2) + C\]

Exam Example: \(\int x^2 \cdot e^{-x} \, dx\)

This exact type appeared on the exam!

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First application: \(u = x^2\), \(dv = e^{-x} \, dx\), so \(v = -e^{-x}\)

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\[\int x^2 \cdot e^{-x} \, dx = -x^2 \cdot e^{-x} - \int -2x \cdot e^{-x} \, dx\]

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\[= -x^2 \cdot e^{-x} + 2\int x \cdot e^{-x} \, dx\]

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Second application: \(u = x\), \(dv = e^{-x} \, dx\), so \(v = -e^{-x}\)

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\[\int x \cdot e^{-x} \, dx = -x \cdot e^{-x} + \int e^{-x} \, dx = -x \cdot e^{-x} - e^{-x}\]

Exam Example: Final Answer

Substituting back:

\[\int x^2 \cdot e^{-x} \, dx = -x^2 \cdot e^{-x} + 2(-x \cdot e^{-x} - e^{-x}) + C\]

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\[= -x^2 \cdot e^{-x} - 2x \cdot e^{-x} - 2e^{-x} + C\]

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\[= -e^{-x}(x^2 + 2x + 2) + C\]

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Verification is important! Differentiate your answer if you have the time: \[\frac{d}{dx}[-e^{-x}(x^2+2x+2)] = e^{-x}(x^2+2x+2-2x-2) = x^2 e^{-x}\]

Part E: Definite Integrals by Parts

Definite Integrals with Bounds

For definite integrals:

\[\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du\]

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Example: \(\int_0^1 x \cdot e^x \, dx\), we found \(\int x \cdot e^x \, dx = e^x(x - 1) + C\)

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\[\int_0^1 x \cdot e^x \, dx = [e^x(x-1)]_0^1 = e^1(1-1) - e^0(0-1)\]

\[= 0 - (-1) = 1\]

Definite Integral Example with Graph

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Essentially, we use integration by parts and then compute the definite integral as learned.

Guided Practice - 20 Minutes

Practice Set A: Single Application

Work individually for 5 minutes

\(\int 3x \cdot e^x \, dx\)
\(\int (2x + 1) \cdot e^x \, dx\)
\(\int x \cdot e^{-x} \, dx\)
\(\int x^2 \cdot \ln(x) \, dx\)

Practice Set B: Repeated Application

Work individually for 7 minutes

\(\int x^2 \cdot e^{2x} \, dx\)
\(\int (x^2 + x) \cdot e^x \, dx\)
\(\int_0^2 x \cdot e^x \, dx\) (evaluate with bounds)

Coffee Break - 15 Minutes

Part F: Average Value of a Function

The Average Value Formula

Key idea: What is the “average height” of a function over an interval?

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\[f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx\]

The Average Visualized

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Makes sense, doesn’t it?

Example: Average Value

Find the average value of \(f(x) = x^2\) on \([0, 4]\)

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\[f_{avg} = \frac{1}{4-0} \int_0^4 x^2 \, dx\]

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\[= \frac{1}{4} \left[\frac{x^3}{3}\right]_0^4\]

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\[= \frac{1}{4} \cdot \frac{64}{3} = \frac{16}{3} \approx 5.33\]

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The area under \(f(x)\) equals the area of a rectangle with height \(f_{avg}\) and width \((b-a)\).

Part G: Revenue and Cost Accumulation

Total Revenue from Marginal Revenue

Key relationship: If \(MR(x)\) is the marginal revenue, then total revenue from selling \(a\) to \(b\) units is:

\[\text{Revenue}_{a \to b} = \int_a^b MR(x) \, dx\]

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Example: \(MR(x) = 100 - 2x\), revenue from selling units 10 to 30:

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\[\int_{10}^{30} (100 - 2x) \, dx = [100x - x^2]_{10}^{30}\]

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\[= (3000 - 900) - (1000 - 100) = 2100 - 900 = 1200 \text{ EUR}\]

Profit Accumulation Over Time

Scenario: Profit rate \(P'(t) = 8t - t^2\) thousand EUR per month (\(t\))

Find total profit from month 2 to month 6.

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\[P_{2 \to 6} = \int_2^6 (8t - t^2) \, dt = \left[4t^2 - \frac{t^3}{3}\right]_2^6\]

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\[= \left(144 - 72\right) - \left(16 - \frac{8}{3}\right)\]

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\[= 72 - \frac{40}{3} = \frac{216 - 40}{3} = \frac{176}{3} \approx 58.67 \text{ thousand EUR}\]

Collaborative Problem-Solving - 25 Minutes

Practice Set C: Synthesis

Work with a partner for 10 minutes

Find the average value of \(f(x) = 2x + 1\) on \([0, 4]\)
A company’s marginal profit is \(P'(x) = 80 - 4x\) EUR/unit.
1. Find the production level \(x^*\) that maximizes profit.
2. Calculate the total profit from \(x = 5\) to \(x = 15\).
A machine’s production rate is \(P(t) = 100e^{-0.1t}\) units/hour. Find the average production rate over the first 10 hours.

Challenge: Present Value Analysis

Scenario: A company expects future profits that decay over time. The profit rate is the follwoing where \(t\) is years and profit is in thousands of euros.

\[P'(t) = (100 + 20t) \cdot e^{-0.05t}\]

Find the antiderivative of the profit rate using integration by parts
Calculate the total profit from year 0 to year 10
At what rate is profit changing at \(t = 5\)?
Interpret the results for a business decision

Wrap-Up & Key Takeaways

What We’ve Learned in Section 06

Antiderivatives reverse differentiation: \(F'(x) = f(x)\)
Definite integrals compute net accumulated change
Area under curves using \(\int_a^b f(x) \, dx\)
Area between curves using \(\int_a^b [f(x) - g(x)] \, dx\)
Consumer Surplus: \(CS = \int_0^{q^*} [D(q) - p^*] \, dq\)
Producer Surplus: \(PS = \int_0^{q^*} [p^* - S(q)] \, dq\)
Integration by Parts: \(\int u \, dv = uv - \int v \, du\)
Average Value: \(f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx\)

Essential Formulas Summary

Concept	Formula
Antiderivative	\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\)
FTC	\(\int_a^b f(x) \, dx = F(b) - F(a)\)
Area Between	\(\int_a^b [f(x) - g(x)] \, dx\)
Consumer Surplus	\(\int_0^{q^} [D(q) - p^] \, dq\)
Producer Surplus	\(\int_0^{q^} [p^ - S(q)] \, dq\)
Integration by Parts	\(\int u \, dv = uv - \int v \, du\)
Average Value	\(\frac{1}{b-a} \int_a^b f(x) \, dx\)

Final Assessment - 5 Minutes

Quick Check

Work individually

Find \(\int (x + 2) \cdot e^x \, dx\)
Find the average value of \(f(x) = x^2\) on \([1, 3]\).
Why does LIATE put logarithms before algebraic functions?
What does consumer surplus measure economically?

Next Session Preview

Session 06-06: Mock Exam

Comprehensive Assessment

Full mock exam covering differential and integral calculus
180 minutes working time with problems covering:
- Graphical differentiation and integration
- Function analysis with calculus
- Area calculations
- Economic applications
- Integration by parts

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Complete tasks and review all integration techniques!