Lesson 4: FTC Part 2 and Net Change Theorem
Estimated time: 35-45 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- State and apply FTC Part 2 to differentiate integrals with variable upper limits
- Use the chain rule with FTC Part 2 for composite upper limits
- State and apply the Net Change Theorem
- Interpret definite integrals as net change in real-world contexts
FTC Part 2: Differentiation of Integrals
While FTC Part 1 lets us evaluate definite integrals, FTC Part 2 tells us how to differentiate an integral whose upper limit is a variable. This reveals integration and differentiation as inverse operations.
Fundamental Theorem of Calculus, Part 2
If f is continuous on an interval containing a, then the function
g(x) = ∫ax f(t) dt
is differentiable, and
g'(x) = f(x)
That is, d/dx [∫ax f(t) dt] = f(x).
What this means: If you define a function by integrating f from a constant to x, then the derivative of that function gives you back the original integrand evaluated at x. Differentiation undoes integration.
Basic FTC Part 2 Examples
Example 1: Direct Application
Find g'(x) if g(x) = ∫1x (t² + 3t) dt.
Solution: By FTC Part 2, simply replace t with x in the integrand:
g'(x) = x² + 3x.
No antiderivative computation needed!
Example 2: Integrand with Special Functions
Find d/dx [∫0x cos(t²) dt].
Solution: Even though cos(t²) has no elementary antiderivative, FTC Part 2 gives us:
d/dx [∫0x cos(t²) dt] = cos(x²).
Example 3: Variable Lower Limit
Find d/dx [∫x5 et² dt].
Solution: First reverse the limits: ∫x5 et² dt = −∫5x et² dt.
Now apply FTC Part 2: d/dx [−∫5x et² dt] = −ex².
FTC Part 2 with the Chain Rule
When the upper limit is not simply x but a function u(x), we must combine FTC Part 2 with the chain rule.
FTC Part 2 + Chain Rule
If g(x) = ∫au(x) f(t) dt, then:
g'(x) = f(u(x)) · u'(x)
Example 4: Composite Upper Limit
Find d/dx [∫0x² sin(t) dt].
Step 1: Here u(x) = x², so u'(x) = 2x.
Step 2: f(t) = sin(t), so f(u(x)) = sin(x²).
Step 3: g'(x) = sin(x²) · 2x = 2x sin(x²).
Example 5: Another Composite Limit
Find d/dx [∫1ln x (t³ + 1) dt].
Step 1: u(x) = ln x, so u'(x) = 1/x.
Step 2: f(t) = t³ + 1, so f(ln x) = (ln x)³ + 1.
Step 3: g'(x) = [(ln x)³ + 1] · (1/x) = [(ln x)³ + 1] / x.
The Net Change Theorem
The Net Change Theorem is a direct consequence of FTC Part 1, reinterpreted as a statement about accumulation.
Net Change Theorem
If F'(x) = f(x) is continuous on [a, b], then:
∫ab F'(x) dx = F(b) − F(a)
The integral of a rate of change gives the net change in the quantity.
Common interpretations:
- If v(t) is velocity, then ∫ab v(t) dt = s(b) − s(a) = net displacement.
- If r(t) is a flow rate, then ∫ab r(t) dt = net amount that flowed.
- If P'(t) is a population growth rate, then ∫ab P'(t) dt = net population change.
Displacement vs. Total Distance
An important distinction arises when the rate of change (like velocity) can be positive or negative.
Displacement vs. Total Distance
Displacement (net change in position): ∫ab v(t) dt
Total distance traveled: ∫ab |v(t)| dt
Example 6: Displacement vs. Distance
A particle moves with velocity v(t) = t² − 4 on [0, 3]. Find the displacement and total distance.
Displacement: ∫03 (t² − 4) dt = [t³/3 − 4t]03 = (9 − 12) − 0 = −3.
The particle ends up 3 units behind where it started.
Total distance: v(t) = 0 when t = 2 (in [0, 3]). v < 0 on [0, 2], v > 0 on [2, 3].
∫02 |t² − 4| dt + ∫23 (t² − 4) dt = ∫02 (4 − t²) dt + ∫23 (t² − 4) dt
= [4t − t³/3]02 + [t³/3 − 4t]23 = (8 − 8/3) + [(9 − 12) − (8/3 − 8)]
= 16/3 + [−3 + 16/3] = 16/3 + 7/3 = 23/3.
Net Change in Applied Contexts
Example 7: Water Flow
Water flows into a tank at a rate of r(t) = 200 − 4t liters per minute. How much water enters during the first 30 minutes?
Solution: Net water = ∫030 (200 − 4t) dt = [200t − 2t²]030 = (6000 − 1800) − 0 = 4200 liters.
Example 8: Population Growth
A bacteria population grows at rate P'(t) = 100e0.1t bacteria per hour. Find the net change in population over [0, 5] hours.
Solution: ∫05 100e0.1t dt = [1000 e0.1t]05 = 1000(e0.5 − 1) ≈ 1000(1.6487 − 1) = 648.7 bacteria.
Key Takeaways
- FTC Part 2: d/dx [∫ax f(t) dt] = f(x). Differentiation undoes integration.
- With a composite upper limit u(x): d/dx [∫au(x) f(t) dt] = f(u(x)) · u'(x).
- Net Change Theorem: ∫ab F'(x) dx = F(b) − F(a). The integral of a rate equals the net change.
- Displacement = ∫ v dt (signed), total distance = ∫ |v| dt (always positive).
Check Your Understanding
1. Find d/dx [∫2x √(1 + t³) dt].
2. Find d/dx [∫0x³ sin(t) dt].
3. A car has velocity v(t) = 3t − 6 m/s on [0, 4]. Find the displacement and total distance.
4. If oil leaks at rate r(t) = 50/(1 + t) gallons/hr, how much oil leaks in the first 4 hours?