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

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Fundamental Theorem of Calculus Part 2: Evaluation

TL;DR

FTC 2 says: if you want to compute a definite integral, find an antiderivative of the integrand, then evaluate it at the upper limit minus the lower limit. This turns integration—which used to mean "find the area under a curve"—into a straightforward algebra problem: no limits, no Riemann sums, just plug and calculate.

1. The Mental Model

The key insight is that differentiation and integration are inverse operations—if you undo a derivative, you get the original function. FTC 2 exploits this: to measure the total accumulation of a quantity (the integral), you don't sum infinite rectangles; instead, you find what function changes at the rate you're given, evaluate that function at your endpoints, and take the difference. It's the bridge from geometry to algebra, and it makes integration actually computable.

2. The Core Material

The Statement

The Fundamental Theorem of Calculus, Part 2 says:

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ on $[a, b]$ (meaning $F'(x) = f(x)$), then:

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

We often write this as $\int_a^b f(x) \, dx = [F(x)]_a^b = F(b) - F(a)$.

That's it. That's the whole theorem. It says: integrate by finding an antiderivative, then subtract.

Why This Works (Intuition Without Rigour)

Imagine you're driving and you know your speed at every moment, $v(t)$. The total distance you've travelled from time $a$ to time $b$ is $\int_a^b v(t) \, dt$. But you could also measure distance as: (position at time $b$) minus (position at time $a$), which is $s(b) - s(a)$. Since $\frac{ds}{dt} = v(t)$, the position function $s(t)$ is an antiderivative of velocity. These two measurements give the same answer—the theorem formalizes that.

How to Use It (The Algorithm)

1. Identify the integrand. This is the function $f(x)$ under the integral sign.
2. Find an antiderivative $F(x)$. Use the power rule in reverse, the reverse of the chain rule (substitution), integration by parts, or tables. Any antiderivative works; they all differ by a constant anyway, and that constant cancels in the subtraction.
3. Evaluate $F$ at the upper limit and lower limit.
4. Subtract: $F(b) - F(a)$.

The Antiderivative vs. the Indefinite Integral

A quick note: $\int f(x) \, dx = F(x) + C$ is the indefinite integral (a family of functions). When you use FTC 2, you pick one antiderivative $F$, then evaluate. The $+C$ doesn't matter because:
$$[F(x) + C]_a^b = [F(b) + C] - [F(a) + C] = F(b) - F(a).$$
The constants cancel. So in practice, just find the antiderivative without worrying about $+C$ when you're computing a definite integral.

3. Worked Example

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

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

Step 2: Find an antiderivative.
- Antiderivative of $3x^2$ is $\frac{3x^3}{3} = x^3$.
- Antiderivative of $-2x$ is $-\frac{2x^2}{2} = -x^2$.
- Antiderivative of $1$ is $x$.

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

Step 3 & 4: Evaluate and subtract.
$$\int_1^3 (3x^2 - 2x + 1) \, dx = [x^3 - x^2 + x]_1^3$$

At $x = 3$: $F(3) = 3^3 - 3^2 + 3 = 27 - 9 + 3 = 21$.

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

$$F(3) - F(1) = 21 - 1 = 20.$$

Answer: $\int_1^3 (3x^2 - 2x + 1) \, dx = 20$.

Another example with a transcendental function:

$\int_0^{\pi/2} \cos(x) \, dx$.

Antiderivative of $\cos(x)$ is $\sin(x)$.

$$[\sin(x)]_0^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1.$$

The area under a cosine hump from $0$ to $\pi/2$ is exactly $1$.

4. Key Takeaways

4.1 Most Important Concepts

1. FTC 2 connects antiderivatives to definite integrals. It says the definite integral equals the antiderivative evaluated at the endpoints.

2. An antiderivative is a function whose derivative is your integrand. You recover $f$ by differentiating $F$.

3. The choice of antiderivative doesn't matter. All antiderivatives differ by a constant, which cancels when you subtract.

4. Integration is now algebraic, not geometric. You no longer compute limits of Riemann sums; you just apply rules to find $F$, then do arithmetic.

5. The integrand must be continuous on the interval. If it jumps or has a vertical asymptote, FTC 2 may not apply (or needs care).

6. Direction matters for the bounds. $\int_a^b f = -\int_b^a f$ because $F(a) - F(b) = -(F(b) - F(a))$.

7. This theorem is the reason we teach antiderivatives. Without FTC 2, the indefinite integral would be an abstract object with little computational use.

4.2 Common Misconceptions

5. Now Try It

Exercise 1 (Polynomial): Compute $\int_0^2 (2x^3 + x) \, dx$.

Show all steps: write the antiderivative $F(x)$, evaluate $F(2)$ and $F(0)$, then report $F(2) - F(0)$.

Exercise 2 (Exponential + Trig): Compute $\int_0^1 (e^x + \sin(x)) \, dx$.

Recall: antiderivative of $e^x$ is $e^x$; antiderivative of $\sin(x)$ is $-\cos(x)$.

Exercise 3 (Check Your Reasoning): Explain why $\int_{-1}^1 x^2 \, dx$ cannot equal $\frac{x^3}{3} |_{-1}^1 - C$ for any non-zero $C$. (Hint: compute both sides.)

Success looks like: You've set up the antiderivative correctly using rules you know, evaluated it at both limits, and computed a single numerical answer for each.

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