Applying FTC
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 1.3: “The Fundamental Theorem of Calculus” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/1-3-the-fundamental-theorem-of-calculus |
| Textbook used in class | Stewart, Calculus, Section 4.3: “The Fundamental Theorem of Calculus” |
Opening Scenario
The Fundamental Theorem of Calculus (FTC) comes in two parts that work in opposite directions. Part 1 says: differentiate an accumulation integral and you get back the integrand. Part 2 says: evaluate a definite integral by finding any antiderivative and subtracting its values at the endpoints.
Together, they collapse the painful limit-of-Riemann-sums calculation into a two-step procedure: find an antiderivative, plug in the bounds, subtract.
Quick Reference
FTC Part 2. If $F$ is any antiderivative of $f$ on $[a, b]$ (meaning $F' = f$), then $$\int_a^b f(x)\,dx = F(b) - F(a).$$
Notation. $F(x)\Big|_a^b = F(b) - F(a)$.
FTC Part 1 with chain rule. If $g$ is differentiable and $f$ is continuous, then $$\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\cdot g'(x).$$
Key Concepts
1. Evaluating Definite Integrals via FTC Part 2
The procedure for $\displaystyle\int_a^b f(x)\,dx$:
- Find any antiderivative $F$ of $f$. (The constant $+C$ cancels in $F(b) - F(a)$, so omit it.)
- Evaluate $F(b) - F(a)$.
Example 1. $\displaystyle\int_1^3 x^2\,dx$.
Antiderivative: $F(x) = \dfrac{x^3}{3}$.
$F(3) - F(1) = \dfrac{27}{3} - \dfrac{1}{3} = 9 - \dfrac{1}{3} = \dfrac{26}{3}$.
Boxed answer: $\dfrac{26}{3}$.
This is far easier than the Riemann sum limit calculation that would otherwise be needed.
2. The Constant Cancels
Any antiderivative works. If you use $F(x) = \dfrac{x^3}{3} + 7$, then $F(3) - F(1) = \left(\dfrac{27}{3} + 7\right) - \left(\dfrac{1}{3} + 7\right) = \dfrac{26}{3}$. The $+7$ cancels. This is why the constant of integration is omitted when evaluating definite integrals.
3. FTC Part 1 with the Chain Rule
FTC Part 1 states that $\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt = f(x)$ when the upper limit is exactly $x$. When the upper limit is a function $g(x)$, the chain rule applies: $$\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\cdot g'(x).$$
Think of it as: differentiate the integrand evaluated at the upper limit, then multiply by the derivative of the upper limit.
Example 2. Find $\dfrac{d}{dx}\displaystyle\int_1^{x^3} \sin(t^2)\,dt$.
Upper limit: $g(x) = x^3$, $g'(x) = 3x^2$.
$\dfrac{d}{dx}\displaystyle\int_1^{x^3} \sin(t^2)\,dt = \sin\!\left((x^3)^2\right)\cdot 3x^2 = 3x^2\sin(x^6)$.
4. Verifying by Differentiating
The FTC means that the derivative of a definite integral (with variable upper limit) should return the integrand. A useful check: if your antiderivative $F$ has $F' = f$, then $F(b) - F(a)$ correctly evaluates $\displaystyle\int_a^b f$.
“You must use the same antiderivative everywhere in a calculation.” Any antiderivative works for FTC Part 2 because the constant cancels. However, you should not mix different antiderivatives within a single problem.
Worked Example
Evaluate $\displaystyle\int_0^{\pi} \cos x\,dx$.
$F(x) = \sin x$, which satisfies $F' = \cos x$.
$\displaystyle\int_0^{\pi} \cos x\,dx = \sin(\pi) - \sin(0) = 0 - 0 = 0$.
Interpretation. The cosine function is positive on $[0, \pi/2]$ and negative on $[\pi/2, \pi]$, and the two areas are equal. The integral confirms this: the signed areas cancel.
Find $\dfrac{d}{dx}\displaystyle\int_x^5 \sqrt{1 + t^4}\,dt$.
Step 1 -- Rewrite with variable in the upper limit. The variable $x$ is in the lower limit, so write: $$\frac{d}{dx}\int_x^5 \sqrt{1+t^4}\,dt = -\frac{d}{dx}\int_5^x \sqrt{1+t^4}\,dt.$$
Step 2 -- Apply FTC Part 1.
$= -\sqrt{1 + x^4}$.
Boxed answer: $-\sqrt{1 + x^4}$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Evaluating $F(a) - F(b)$ instead of $F(b) - F(a)$ | Writing $\int_1^3 x\,dx = F(1) - F(3)$ | Upper bound minus lower bound: $F(b) - F(a)$ |
| Forgetting the chain rule in FTC Part 1 | $\frac{d}{dx}\int_0^{x^2} f(t)\,dt = f(x^2)$ | Must multiply by $(x^2)' = 2x$: the answer is $2x\,f(x^2)$ |
| Including $+C$ when using FTC Part 2 | Writing $F(b) - F(a) + C$ | The constants cancel; omit $C$ for definite integrals |
Leveled Practice
Level 1 -- FTC Part 2
Problem 1. Evaluate $\displaystyle\int_1^4 (3x^2 - 2x + 1)\,dx$.
Show answer
Antiderivative: $F(x) = x^3 - x^2 + x$.
$F(4) - F(1) = (64 - 16 + 4) - (1 - 1 + 1) = 52 - 1 = 51$.
Boxed answer: $51$.
Level 2 -- FTC Part 2 with Trig
Problem 2. Evaluate $\displaystyle\int_0^{\pi/2}(\sin x + \cos x)\,dx$.
Show answer
Antiderivative: $F(x) = -\cos x + \sin x$.
$F(\pi/2) - F(0) = (-\cos(\pi/2) + \sin(\pi/2)) - (-\cos 0 + \sin 0)$ $= (0 + 1) - (-1 + 0) = 1 + 1 = 2$.
Boxed answer: $2$.
Level 3 -- FTC Part 1 with Chain Rule
Problem 3. Find $\dfrac{d}{dx}\displaystyle\int_2^{x^2} e^{t^3}\,dt$.
Show answer
Upper limit $g(x) = x^2$, $g'(x) = 2x$.
$\dfrac{d}{dx}\displaystyle\int_2^{x^2} e^{t^3}\,dt = e^{(x^2)^3}\cdot 2x = 2x\,e^{x^6}$.
Mastery Checklist
Mental Model
FTC Part 2 says: the net change in $F$ from $a$ to $b$ (namely $F(b) - F(a)$) equals the integral of $F$’s rate of change ($f = F'$) over $[a, b]$. This is “net change = integral of rate of change,” the fundamental accounting identity of calculus. The limit-of-Riemann-sums definition gives the same answer because it directly accumulates infinitesimal changes $f\,dx = dF$ along the way.
Connections
Looking back
- Antiderivatives (Section 3.9): FTC Part 2 uses antiderivatives computed in Section 3.9 and 4.4.
- Definite integral (Section 4.2): FTC Part 2 provides the practical evaluation tool for the integral defined in 4.2.
Looking ahead
- Indefinite integrals (Section 4.4): The antiderivative notation $\int f(x)\,dx = F(x) + C$ is directly tied to FTC Part 2.
- Substitution (Section 4.5): Extends FTC Part 2 to more complex integrands by first simplifying via a change of variable.
Back to Integration Foundations | Next: Indefinite Integrals