Fundamental Theorem of Calculus Part 2 (FTC 2): Evaluation

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Fundamental Theorem of Calculus Part 2: Evaluation

TL;DR

FTC Part 2 says: if you want to compute a definite integral, find an antiderivative F of your integrand f, then the answer is just F(b) − F(a). This turns a hard limiting process into arithmetic. It's the bridge between derivatives (rates) and integrals (accumulation).

1. The Mental Model

Think of the definite integral ∫ₐᵇ f(x) dx as "how much stuff accumulates from a to b?" To answer it, you reverse-engineer: find a function F whose derivative is f. Then the total accumulation is simply "final value minus starting value." The integral isn't computing area by slicing anymore—it's just evaluating a function at two points.

2. The Core Material

What FTC Part 2 Actually Says

The Fundamental Theorem of Calculus, Part 2 states:

If f is continuous on [a, b] and F is any antiderivative of f on [a, b], then:

$$\int_a^b f(x) \, dx = F(b) - F(a)$$

We write this with bracket notation: $\left[ F(x) \right]_a^b = F(b) - F(a)$

Why does this work? Because of the First Fundamental Theorem: the derivative of an accumulation function is the original function. FTC Part 2 flips that—if you have a function whose derivative is f, you can "undo" the derivative to evaluate the integral. You're not computing area by summing up infinitely many rectangles. You're just plugging in two numbers.

Finding an Antiderivative (The Essential Skill)

You already know how to find antiderivatives from integration rules:
- $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$)
- $\int e^x \, dx = e^x + C$
- $\int \sin(x) \, dx = -\cos(x) + C$
- $\int \frac{1}{x} \, dx = \ln|x| + C$

Key point: When you use FTC Part 2, you don't need the "+ C". The constant cancels:
$$F(b) - F(a) = (f(x) + C)|_{x=b} - (f(x) + C)|_{x=a} = f(b) - f(a) + C - C = f(b) - f(a)$$

So you find any antiderivative—pick C = 0 for simplicity—then evaluate at the bounds.

Three Steps to Solve Any Definite Integral

1. Recognise the integrand. What function is under the integral? Is it a polynomial, exponential, trig, or a combo?
2. Find an antiderivative F(x). Use your integration rules. Don't worry about "+ C".
3. Evaluate at the bounds. Plug in the upper limit, subtract the value at the lower limit.

# Example: definite integral of 2x from 0 to 3
# Step 1: f(x) = 2x
# Step 2: F(x) = x^2  (antiderivative; ignoring +C)
# Step 3: F(3) - F(0) = 9 - 0 = 9

def antiderivative(x):
    return x**2

result = antiderivative(3) - antiderivative(0)
print(result)  # Output: 9

Why This Matters: Escaping the Limit Definition

Without FTC Part 2, every definite integral forces you back to Riemann sums:
$$\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x$$

That's brutal for anything beyond tiny intervals. FTC Part 2 obliterates this: one antiderivative, two function evaluations, done. This is why calculus became useful in physics and engineering—not because of the abstract limit, but because we can compute integrals in seconds.

3. Worked Example

Problem: Compute $\int_1^4 (3x^2 - 2x + 1) \, dx$

Step 1: Identify f(x)
$$f(x) = 3x^2 - 2x + 1$$

Step 2: Find an antiderivative F(x)
Integrate term by term:
- $\int 3x^2 \, dx = x^3$
- $\int -2x \, dx = -x^2$
- $\int 1 \, dx = x$

So $F(x) = x^3 - x^2 + x$

Step 3: Evaluate at the bounds
$$\left[ x^3 - x^2 + x \right]_1^4$$

At $x = 4$:
$$F(4) = 4^3 - 4^2 + 4 = 64 - 16 + 4 = 52$$

At $x = 1$:
$$F(1) = 1^3 - 1^2 + 1 = 1 - 1 + 1 = 1$$

Answer:
$$\int_1^4 (3x^2 - 2x + 1) \, dx = 52 - 1 = 51$$

Verify mentally: the integrand is an upward-opening cubic. It's positive over [1, 4], and grows fast. An answer of 51 square units feels reasonable.

4. Key Takeaways

4.1 Most Important Concepts

1. FTC Part 2 is an evaluation shortcut. Instead of computing limits of Riemann sums, find F' = f, then subtract endpoints.

2. Antiderivatives are the key. The whole theorem rests on your ability to reverse differentiation. Master your integration rules first.

3. The constant cancels automatically. You don't need to track "+ C" in definite integrals—it vanishes when you subtract.

4. Continuity is required. If f has a jump discontinuity inside [a, b], FTC Part 2 breaks. The theorem demands f is continuous on the closed interval.

5. Definite vs. indefinite are linked. A definite integral ∫ₐᵇ f(x) dx is just an indefinite integral F(x) evaluated at two points.

6. Direction matters. $\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx$. Swapping limits flips the sign.

4.2 Common Misconceptions

4.3 Compare & Contrast

5. Now Try It

Exercise: Compute $\int_0^2 (e^x + 4x^3) \, dx$

Work through all three steps:
1. Write down f(x).
2. Find F(x). (Hint: the antiderivative of e^x is e^x, and the antiderivative of 4x^3 is x^4.)
3. Evaluate [F(x)] from 0 to 2, then compute F(2) − F(0).

Once you have a numerical answer, check it makes sense: e^x is always positive, and 4x^3 is positive on [0, 2], so the result should be positive. Also: e^2 ≈ 7.4, so the exponential part alone contributes roughly 7.4 − 1 ≈ 6.4. The polynomial part adds on top. Your answer should be more than 6.

Success looks like: You write down the antiderivative correctly, substitute both bounds, subtract, and get a single number (approximately 16.4).

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