
Section 06: Integral Calculus
Test your understanding of Definite Integrals
Evaluate \(\int_1^4 (3x^2 - 2x) \, dx\)
If \(\int_0^5 f(x) \, dx = 18\) and \(\int_3^5 f(x) \, dx = 7\), find \(\int_0^3 f(x) \, dx\)
What does \(\int_a^b f'(x) \, dx\) represent geometrically and algebraically?
For \(f(x) = x - 2\) on \([0, 4]\), is the signed area positive, negative, or zero?
Focus on FTC and definite integrals
Today we focus on area calculations and introduce exponential and logarithmic integrals!
Area calculations are one of the most common applications of integration!
Simple case: When \(f(x) \geq 0\) on \([a, b]\), the definite integral gives the area directly.
\[\text{Area} = \int_a^b f(x) \, dx\]

Find the area under \(f(x) = x^2\) from \(x = 0\) to \(x = 3\).
Solution:
\[\text{Area} = \int_0^3 x^2 \, dx = \frac{x^3}{3} \Big|_0^3 = \frac{27}{3} - 0 = 9\]
Find the area under \(f(x) = x^2\) from \(x = 0\) to \(x = 3\).

When \(f(x) < 0\): The definite integral gives a negative value!

Definite integral ≠ Total area when function crosses the x-axis!
To find total (unsigned) area:
\[\text{Total Area} = \sum |\text{each region}|\]
Total area between \(f(x) = x^2 - 4\) and x-axis from \(x = 0\) to \(x = 3\)?
Step 1: Find zeros: \(x^2 - 4 = 0 \implies x = \pm 2\)
Step 2: Determine signs:
Step 3: Calculate each piece:
\[\int_0^2 (x^2 - 4) \, dx = \left[\frac{x^3}{3} - 4x\right]_0^2 = \frac{8}{3} - 8 = -\frac{16}{3}\]
\[\int_2^3 (x^2 - 4) \, dx = \left[\frac{x^3}{3} - 4x\right]_2^3 = (9 - 12) - (\frac{8}{3} - 8) = -3 + \frac{16}{3} = \frac{7}{3}\]
Step 4: Total area:
\[\text{Total Area} = \left|-\frac{16}{3}\right| + \frac{7}{3} = \frac{16}{3} + \frac{7}{3} = \frac{23}{3}\]

Total Area = \(\frac{16}{3} + \frac{7}{3} = \frac{23}{3} \approx 7.67\)
Recall: \(\frac{d}{dx}[e^x] = e^x\)
Therefore:
\[\int e^x \, dx = e^x + C\]
The exponential function is its own antiderivative!
For \(e^{ax}\) where \(a\) is a constant:
\[\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C\]
Verification: \(\frac{d}{dx}\left[\frac{1}{a}e^{ax}\right] = \frac{1}{a} \cdot a \cdot e^{ax} = e^{ax}\) ✓
Trick: Divide by the coefficient of \(x\) in the exponent.
Example 1: \(\int e^{3x} \, dx\)
Example 2: \(\int e^{-2x} \, dx\)
Example 3: \(\int 4e^{5x} \, dx\)
Recall the power rule:
\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad \text{for } n \neq -1\]
Why not \(n = -1\)? Division by zero!
\[\int x^{-1} \, dx = \int \frac{1}{x} \, dx = \text{???}\]
\[\int \frac{1}{x} \, dx = \ln|x| + C\]
\[\int \frac{1}{x} \, dx = \ln|x| + C\]
The absolute value handles both positive and negative \(x\).
Example 1: \(\int_1^e \frac{1}{x} \, dx\)
Example 2: \(\int_1^4 \frac{3}{x} \, dx\)
Example 3: \(\int_{-3}^{-1} \frac{2}{x} \, dx\)
| Function | Antiderivative |
|---|---|
| \(e^x\) | \(e^x + C\) |
| \(e^{ax}\) | \(\frac{1}{a}e^{ax} + C\) |
| \(\frac{1}{x}\) | \(\ln|x| + C\) |
| \(\frac{1}{ax+b}\) | \(\frac{1}{a}\ln|ax+b| + C\) |
These formulas will appear frequently in business applications!
Work individually for 8 minutes
Find the area under \(f(x) = 3x^2\) from \(x = 1\) to \(x = 4\).
Find the total area between \(f(x) = x - 3\) and the x-axis from \(x = 0\) to \(x = 5\).
Find the total area between \(f(x) = x^2 - 1\) and the x-axis from \(x = 0\) to \(x = 2\).
Work individually for 6 minutes
\(\int e^{4x} \, dx\)
\(\int 5e^{-x} \, dx\)
\(\int_0^2 e^{3x} \, dx\)
\(\int_1^5 \frac{2}{x} \, dx\)
\(\int (e^x + \frac{1}{x}) \, dx\)
Work in pairs for 6 minutes
Find the area enclosed between \(f(x) = e^x\) and the x-axis from \(x = 0\) to \(x = 2\).
A population decays according to \(P(t) = 1000e^{-0.1t}\). Find the average population from \(t = 0\) to \(t = 10\). (Hint: Average = \(\frac{1}{b-a}\int_a^b f(x) \, dx\))
Scenario: A company’s profit rate (profit per month) is:
\[P'(t) = 50 - 2t \text{ thousand euros per month}\]
where \(t\) is months since launch.
Questions:
Part 1: Total profit
\[\int_0^{12} (50 - 2t) \, dt = [50t - t^2]_0^{12} = 600 - 144 = 456\]
Total profit = €456,000
Part 2: When profit rate becomes zero
\[50 - 2t = 0 \implies t = 25 \text{ months}\]
The profit rate stays positive for the first 25 months.

Scenario: Sales of a product decline exponentially after its peak:
\[S(t) = 10000 \cdot e^{-0.2t} \text{ units per month}\]
Question: What are the total sales from \(t = 0\) to \(t = 6\) months?
Solution:
\[\int_0^6 10000e^{-0.2t} \, dt = 10000 \cdot \frac{1}{-0.2}e^{-0.2t}\Big|_0^6\]
\[= -50000(e^{-1.2} - e^0) = -50000(0.301 - 1) = 34,950 \text{ units}\]
Scenario: An e-commerce company tracks its daily revenue rate:
\[R'(t) = 5000 + 200t - 5t^2 \text{ euros per day}\]
where \(t\) is days since a marketing campaign started.
The campaign runs for 30 days.
Work in groups of 3-4
Graph \(R'(t)\) for the 30-day period. When is the revenue rate highest?
Calculate the total revenue for the first 10 days.
Calculate the total revenue for the entire 30-day campaign.
On which day does the revenue rate first drop below €4,000/day?
Find the average daily revenue rate over the 30-day campaign.
If the campaign costs €80,000, what is the net profit?
Next session: Area between TWO curves and economic surplus!
Work individually, then compare
Find the total area between \(f(x) = x - 2\) and the x-axis from \(x = 0\) to \(x = 4\).
Evaluate \(\int_0^3 2e^{-x} \, dx\).
A company’s revenue rate is \(R'(t) = 100 + 20t\) thousand euros per month. Find total revenue for months 1-5.
Complete Tasks 06-03
Session 06-03 - Area Problems & Basic Applications | Dr. Nikolai Heinrichs & Dr. Tobias Vlćek | Home