FTC Part 1
The Fundamental Theorem of Calculus, Part 1
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 5.3: “The Fundamental Theorem of Calculus” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/5-3-the-fundamental-theorem-of-calculus |
| Supplementary | OpenStax Calculus Volume 1, Section 5.4: “Integration Formulas and the Net Change Theorem” |
| Supplementary link | https://openstax.org/books/calculus-volume-1/pages/5-4-integration-formulas-and-the-net-change-theorem |
| Textbook used in class | Stewart, Calculus, Section 4.3: “The Fundamental Theorem of Calculus” (Examples 1, 2, 3, 4) |
Both OpenStax sources are free and openly licensed.
Key idea
The Fundamental Theorem of Calculus, Part 1, says that integration and differentiation are inverse operations. One undoes the other. That single fact is the reason calculus works as a unified subject, and it is genuinely surprising the first time you meet it.
Here is the setup. Fix a continuous function $f$ and a starting point $a$. Define a new function by accumulating area: \[ g(x) = \int_a^x f(t)\,dt. \] Think of $g(x)$ as the “area so far”: the signed area under $f$ from the start $a$ up to the moving right edge $x$. As $x$ slides to the right, $g$ records how much area has piled up.
Now ask the natural question: how fast is the accumulated area growing? The rate of growth is the derivative $g'(x)$. The theorem’s answer is as clean as it could possibly be: \[ g'(x) = f(x). \] The rate at which area accumulates equals the height of the curve at the right edge. That makes sense from the picture: when you nudge the right edge a little, the sliver of new area is roughly a thin rectangle of height $f(x)$, so area grows at rate $f(x)$.
The consequence is profound. Differentiating an integral gives back the original function. Integration builds up $f$ into an area; differentiation tears that area back down to $f$. They are opposite processes, and recognizing that is what lets you evaluate integrals (in Part 2) without ever summing rectangles again.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
If the difference between the dummy variable $t$ (inside the integral) and the limit variable $x$ (the upper bound) is unclear, slow down on it. The whole theorem is about a function of $x$ built by integrating in $t$.
Quick Reference
The accumulation function. For $f$ continuous on $[a, b]$, define \[ g(x) = \int_a^x f(t)\,dt, \qquad a \leq x \leq b. \] This is a function of $x$; the variable $t$ is just a placeholder inside the integral.
The Fundamental Theorem of Calculus, Part 1. If $f$ is continuous on $[a, b]$, then $g(x) = \int_a^x f(t)\,dt$ is continuous on $[a, b]$, differentiable on $(a, b)$, and \[ g'(x) = f(x). \] In Leibniz notation, \[ \frac{d}{dx}\int_a^x f(t)\,dt = f(x). \]
Plain reading. Differentiating an accumulation function returns the integrand evaluated at the upper limit. Integration and differentiation are inverse processes.
The chain-rule extension. When the upper limit is a function $u(x)$, \[ \frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f\big(u(x)\big)\cdot u'(x). \]
The lower-limit rule. A variable in the lower limit flips the sign: $\dfrac{d}{dx}\int_x^b f(t)\,dt = -f(x)$, because $\int_x^b = -\int_b^x$.
Key Concepts
1. The Accumulation Function
Before the theorem, get comfortable with the object it is about: a function defined by an integral with a moving upper limit.
Fix continuous $f$ and a start $a$, and define $g(x) = \int_a^x f(t)\,dt$. For each fixed $x$, the integral is a definite number, the signed area from $a$ to $x$. Letting $x$ move turns that number into a function of $x$. Think of $g$ as the “area so far.”
Example 1. Let $g(x) = \int_0^x f(t)\,dt$ for the function $f$ whose graph is a triangle of height $2$ over $[0,1]$, then constant. Read off $g(0)$, $g(1)$, $g(2)$ by accumulating area. (This is the read-off computation from Stewart 4.3, Example 1.)
- $g(0) = \int_0^0 f(t)\,dt = 0$: no width, no area.
- $g(1) = \int_0^1 f(t)\,dt = \frac12(1)(2) = 1$: the area of the triangle.
- $g(2) = g(1) + \int_1^2 f(t)\,dt = 1 + (1)(2) = 3$: add the rectangle from $1$ to $2$.
Recap. Accumulation is additive: to extend the area to a new right edge, add the new piece to what you already had. Where $f$ is positive, $g$ increases; where $f$ is negative, $g$ decreases (you subtract area). That rising-and-falling behavior of $g$ is the first hint that $g'$ is controlled by the sign of $f$.
2. The Theorem: Differentiating the Area Function
The central claim is that the accumulation function’s derivative is the original integrand.
The Fundamental Theorem of Calculus, Part 1. If $f$ is continuous on $[a, b]$, then $g(x) = \int_a^x f(t)\,dt$ is differentiable on $(a, b)$ and \[ g'(x) = f(x). \] (Plain gloss: the instantaneous rate at which area accumulates equals the height of the curve at the moving edge.)
The reason, in one picture: increasing the upper limit from $x$ to $x + h$ adds a thin strip of area. For small $h$ that strip is almost a rectangle of width $h$ and height $f(x)$, so \[ g(x + h) - g(x) \approx h\,f(x), \qquad \frac{g(x+h) - g(x)}{h} \approx f(x). \] Letting $h \to 0$ turns the approximation into equality, $g'(x) = f(x)$. (The rigorous version traps the difference quotient between the minimum and maximum of $f$ on the strip and applies the Squeeze Theorem; continuity of $f$ is what makes both bounds converge to $f(x)$.)
A first sanity check from a function you can integrate directly: with $f(t) = t$ and $a = 0$, \[ g(x) = \int_0^x t\,dt = \frac{x^2}{2}, \qquad g'(x) = x = f(x). \] The derivative of the area function is indeed the integrand.
3. Computing Derivatives of Integrals
In practice the theorem is a one-step rule: to differentiate $\int_a^x f(t)\,dt$, just evaluate $f$ at $x$. No integration required.
Example 2. Find the derivative of $g(x) = \int_0^x \sqrt{1 + t^2}\,dt$. (This is Stewart 4.3, Example 2.)
Goal. Apply Part 1 directly: replace $t$ with $x$ in the integrand.
The integrand $f(t) = \sqrt{1 + t^2}$ is continuous, so Part 1 applies: \[ g'(x) = \sqrt{1 + x^2}. \]
Boxed answer: $g'(x) = \sqrt{1 + x^2}$.
Recap. This is striking. The integral $\int_0^x \sqrt{1+t^2}\,dt$ has no elementary closed form, so you could not evaluate it and then differentiate. Yet its derivative is immediate. The theorem lets you differentiate a function you cannot integrate.
4. The Chain-Rule Extension
When the upper limit is a function of $x$ rather than $x$ itself, combine Part 1 with the chain rule.
Example 3. Find $\dfrac{d}{dx}\int_1^{x^4}\sec t\,dt$. (This is Stewart 4.3, Example 4.)
Goal. Let $u = x^4$ be the upper limit. Differentiate the integral with respect to $u$ (Part 1), then multiply by $\dfrac{du}{dx}$ (chain rule).
Let $u = x^4$, so $\dfrac{du}{dx} = 4x^3$. Then \[ \frac{d}{dx}\int_1^{x^4}\sec t\,dt = \frac{d}{du}\left[\int_1^{u}\sec t\,dt\right]\cdot\frac{du}{dx} = \sec u \cdot 4x^3 = 4x^3\sec(x^4). \]
Boxed answer: $\dfrac{d}{dx}\int_1^{x^4}\sec t\,dt = 4x^3\sec(x^4)$.
Recap. The rule is: evaluate the integrand at the upper limit, then multiply by the derivative of that upper limit. Part 1 alone handles the case where the upper limit is plain $x$ (whose derivative is $1$); the chain rule handles everything else.
Important: a variable in the lower limit flips the sign. To differentiate $\int_x^b f(t)\,dt$, first swap the limits: $\int_x^b f(t)\,dt = -\int_b^x f(t)\,dt$. Then Part 1 gives $\dfrac{d}{dx}\int_x^b f(t)\,dt = -f(x)$. Forgetting the sign is the most common error when the variable sits at the bottom of the integral.
5. Integration and Differentiation Are Inverses
The deep content of Part 1 is the relationship it certifies between the two halves of calculus.
Part 1 says that if you integrate $f$ to build the area function $g$, then differentiate $g$, you return to $f$: \[ \frac{d}{dx}\int_a^x f(t)\,dt = f(x). \] Integration accumulates; differentiation reads off the rate of accumulation; the second undoes the first. This is exactly why every continuous function has an antiderivative: the accumulation function $g(x) = \int_a^x f(t)\,dt$ is one, guaranteed to exist even when no formula for it does (as in Example 2).
That existence guarantee is the bridge to Part 2. Because $g$ is an antiderivative of $f$, and because any two antiderivatives differ by a constant, knowing any antiderivative $F$ lets you evaluate a definite integral by subtraction, $\int_a^b f(x)\,dx = F(b) - F(a)$, with no rectangles to sum. Part 1 builds the antiderivative; Part 2 cashes it in.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Trying to integrate first | computing $\int_0^x \sqrt{1+t^2}\,dt$ then differentiating | Part 1 is one step: $g'(x) = \sqrt{1+x^2}$ directly |
| Forgetting the chain rule | $\frac{d}{dx}\int_1^{x^4}\sec t\,dt = \sec(x^4)$ | Multiply by the derivative of the upper limit: $4x^3\sec(x^4)$ |
| Missing the sign on a lower limit | $\frac{d}{dx}\int_x^b f(t)\,dt = f(x)$ | A variable lower limit flips the sign: it is $-f(x)$ |
| Confusing the dummy variable with the limit | writing $g'(x)$ in terms of $t$ | The answer is in terms of $x$; $t$ disappears after differentiating |
| Ignoring the continuity hypothesis | applying Part 1 across a discontinuity of $f$ | The theorem requires $f$ continuous on the interval |
| Thinking $g(a) \neq 0$ | treating the starting value as unknown | $g(a) = \int_a^a f(t)\,dt = 0$ always |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Find $g'(x)$ for $g(x) = \int_1^x \dfrac{1}{1 + t^2}\,dt$.
Show answer
The integrand $f(t) = \dfrac{1}{1+t^2}$ is continuous, and the upper limit is plain $x$. By Part 1, evaluate the integrand at $x$: \[ g'(x) = \frac{1}{1 + x^2}. \]
Boxed answer: $g'(x) = \dfrac{1}{1 + x^2}$. (This is Stewart 4.3, Exercise 9.)
Problem 2. Find $g'(x)$ for $g(x) = \int_0^x (2 + \sin t)\,dt$.
Show answer
The integrand is continuous and the upper limit is $x$. By Part 1: \[ g'(x) = 2 + \sin x. \]
Boxed answer: $g'(x) = 2 + \sin x$. (This is Stewart 4.3, Exercise 8. You can check it: $g(x) = 2x - \cos x + 1$, and $g'(x) = 2 + \sin x$.)
Problem 3. For $g(x) = \int_0^x t^2\,dt$, find $g'(x)$ two ways: by Part 1, and by evaluating the integral first and then differentiating.
Show answer
By Part 1: $g'(x) = x^2$.
By integrating first: $g(x) = \int_0^x t^2\,dt = \dfrac{x^3}{3}$, so $g'(x) = \dfrac{3x^2}{3} = x^2$.
Both give the same answer.
Boxed answer: $g'(x) = x^2$. (This is Stewart 4.3, Exercise 7, and it confirms Part 1 directly.)
Level 2 -- Multiple Steps
Problem 4. Find $\dfrac{d}{dx}\int_0^{x^2}\sin t\,dt$.
Show answer
The upper limit is $u = x^2$, with $\dfrac{du}{dx} = 2x$. By Part 1 plus the chain rule: \[ \frac{d}{dx}\int_0^{x^2}\sin t\,dt = \sin(x^2)\cdot 2x = 2x\sin(x^2). \]
Boxed answer: $2x\sin(x^2)$. (This is Stewart 4.3, Exercise 60.)
Problem 5. Find $\dfrac{dy}{dx}$ for $y = \int_x^{\pi}\cos(t^2)\,dt$.
Show answer
The variable is in the lower limit, so flip the limits first: \[ y = \int_x^{\pi}\cos(t^2)\,dt = -\int_{\pi}^{x}\cos(t^2)\,dt. \] Now Part 1 applies to the upper limit $x$: \[ \frac{dy}{dx} = -\cos(x^2). \]
Boxed answer: $\dfrac{dy}{dx} = -\cos(x^2)$. (This is the form of Stewart 4.3, Exercise 10. The leading minus sign is the key step.)
Problem 6. The Fresnel function $S(x) = \int_0^x \sin\!\left(\dfrac{\pi t^2}{2}\right)dt$ is used in optics and highway design. On what intervals is $S$ increasing?
Show answer
By Part 1, $S'(x) = \sin\!\left(\dfrac{\pi x^2}{2}\right)$. A function increases where its derivative is positive, so $S$ increases where \[ \sin\!\left(\frac{\pi x^2}{2}\right) > 0, \] that is, where $\dfrac{\pi x^2}{2}$ lies in an interval $(2k\pi, (2k+1)\pi)$ for an integer $k \geq 0$. For $x > 0$ this first happens when $0 < \dfrac{\pi x^2}{2} < \pi$, i.e. $0 < x < \sqrt{2}$.
Boxed answer: $S$ is increasing on $(0, \sqrt{2})$ (and on the further intervals where $\sin(\pi x^2/2) > 0$). (This is the analysis behind Stewart 4.3, Example 3 and Exercise 69. Part 1 turns a question about an unintegrable function into a question about the sign of its known derivative.)
Level 3 -- Deeper Problems
Problem 7. Find $\dfrac{d}{dx}\int_{x^2}^{1}\dfrac{1}{1 + t^3}\,dt$.
Show answer
The variable is in the lower limit, so flip first, then apply the chain rule with $u = x^2$: \[ \int_{x^2}^{1}\frac{1}{1 + t^3}\,dt = -\int_{1}^{x^2}\frac{1}{1 + t^3}\,dt. \] \[ \frac{d}{dx}\left(-\int_{1}^{x^2}\frac{1}{1 + t^3}\,dt\right) = -\frac{1}{1 + (x^2)^3}\cdot 2x = -\frac{2x}{1 + x^6}. \]
Boxed answer: $-\dfrac{2x}{1 + x^6}$. (This is Stewart 4.3, Exercise 17. Two effects combine: the sign flip from the lower limit and the $2x$ from the chain rule.)
Problem 8. Let $F(x) = \int_{\pi}^{x}\dfrac{\cos t}{t}\,dt$. Find an equation of the tangent line to $y = F(x)$ at the point where $x = \pi$.
Show answer
The tangent line needs a point and a slope.
Point. At $x = \pi$, $F(\pi) = \int_{\pi}^{\pi}\dfrac{\cos t}{t}\,dt = 0$. The point is $(\pi, 0)$.
Slope. By Part 1, $F'(x) = \dfrac{\cos x}{x}$, so \[ F'(\pi) = \frac{\cos \pi}{\pi} = \frac{-1}{\pi}. \]
Line. Using point-slope form, \[ y - 0 = -\frac{1}{\pi}(x - \pi), \quad \text{or} \quad y = -\frac{1}{\pi}(x - \pi). \]
Boxed answer: $y = -\dfrac{1}{\pi}(x - \pi)$. (This is Stewart 4.3, Exercise 63. The value $F(\pi) = 0$ comes from the integral over a zero-width interval, and the slope comes straight from Part 1.)
Problem 9. Explain what is wrong with the following calculation, and why Part 1 (and Part 2) does not rescue it: \[ \int_{-1}^{3}\frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_{-1}^{3} = -\frac{1}{3} - 1 = -\frac{4}{3}. \]
Show answer
The answer is negative, but the integrand $\dfrac{1}{x^2} \geq 0$, so any genuine area would be nonnegative. Something is wrong.
The error is applying the Fundamental Theorem across a discontinuity. The function $f(x) = \dfrac{1}{x^2}$ has an infinite discontinuity at $x = 0$, which lies inside $[-1, 3]$. The theorem requires $f$ to be continuous on the whole interval, so neither part applies here.
Conclusion: the calculation is invalid because $\frac{1}{x^2}$ is not continuous on $[-1, 3]$; in fact this integral does not exist as an ordinary integral. (This is Stewart 4.3, Example 8.) Always confirm the integrand is continuous on the entire interval before using the Fundamental Theorem.
Common Misconceptions
the derivative of $\int_a^x f(t)\,dt$ is $f(t)$, with the dummy variable $t$ surviving in the answer.
This is the action-view-of-function error. Once the upper limit is set to $x$, the integral is a function of $x$ only; the placeholder $t$ disappears. FTC Part 1 evaluates the integrand at the upper limit: $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$. Retaining $t$ in the answer conflates the integration variable with the limit variable, contradicting the meaning of a definite integral.
differentiating $\int_a^{u(x)} f(t)\,dt$ gives $f(u(x))$ without multiplying by $u'(x)$.
This is the composition-is-not-chaining error. The accumulation function $G(u) = \int_a^u f(t)\,dt$ satisfies $G'(u) = f(u)$, and $h(x) = G(u(x))$ is a composite. The chain rule requires $h'(x) = G'(u(x)) \cdot u'(x) = f(u(x))\cdot u'(x)$. Skipping $u'(x)$ is the same mistake as differentiating $\sin(x^2)$ and forgetting the factor of $2x$.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Think of Part 1 as an odometer reading off a speedometer.
Let $f$ be the speed of a car at each moment. The accumulation function $g(x) = \int_a^x f(t)\,dt$ is the distance traveled by time $x$, the odometer reading. Part 1 says the derivative of the odometer is the speedometer: the rate at which distance accumulates is the current speed, $g'(x) = f(x)$.
Three things follow from the picture:
- Accumulating and rate-reading are opposite moves. Integrate the speed to get distance; differentiate the distance to recover the speed. Part 1 certifies that these undo each other, which is the entire reason calculus is one subject rather than two.
- You can differentiate things you cannot integrate. The accumulation function may have no formula (the integral of $\sqrt{1+t^2}$ has no elementary antiderivative), yet its derivative is instant: the integrand at the upper limit. The theorem sidesteps the integration entirely.
- Every continuous function has an antiderivative, even a nameless one. The “area so far” function is always an antiderivative of $f$, guaranteed to exist by Part 1 whether or not you can write it down. That existence is exactly what Part 2 needs to turn integration into subtraction.
Connections
Within Integrals (Chapter 4)
- Riemann sums and the definite integral: The accumulation function is built from definite integrals, which are limits of Riemann sums. Part 1 is the payoff that frees you from computing those limits by hand.
- The antiderivative concept: Part 1 proves that the accumulation function is an antiderivative of $f$, which is why antiderivatives always exist for continuous functions. The notion of antiderivative becomes concrete here.
- FTC Part 2 and evaluating definite integrals: Part 1 supplies the antiderivative that Part 2 uses to evaluate $\int_a^b f(x)\,dx = F(b) - F(a)$. Together the two parts are the engine of integral calculus, turning area problems into subtraction.
Toward later calculus (MATH161 and beyond)
- Special functions defined by integrals: Many important functions (the Fresnel function, the error function, the sine integral) have no elementary formula and are defined exactly as accumulation functions. Part 1 is what lets you analyze them, since their derivatives are known even when their values are not.
- The net change theorem: Reading Part 2 in reverse, the integral of a rate of change gives the total change, $\int_a^b F'(x)\,dx = F(b) - F(a)$. This turns the abstract theorem into the practical principle that integrating a rate recovers the accumulated quantity.
- Differential equations: Solving $y' = f(x)$ with an initial condition is exactly building an accumulation function. Part 1 guarantees the solution exists and is given by an integral, the starting point for the entire theory of differential equations.
Audience Notes
For students who find math intimidating: The hard-looking part, the integral with $x$ on top, hides an easy rule. To differentiate $\int_a^x f(t)\,dt$, just write $f$ with $x$ plugged in, and stop. If the top is something fancier than $x$, like $x^4$, also multiply by its derivative. If the variable is on the bottom instead of the top, put a minus sign in front. Those three moves cover every problem in this section.
For career-focused students: The odometer-and-speedometer relationship is everywhere in engineering and data work: position is the integral of velocity, energy is the integral of power, total cost is the integral of marginal cost. Part 1 is the formal statement that these accumulated quantities have the original rate as their derivative, which is the backbone of how sensors and simulations relate rates to totals.
For gifted and curious students: Investigate why the accumulation function smooths out a rough integrand: even if $f$ has corners, $g(x) = \int_a^x f(t)\,dt$ is differentiable (its derivative is $f$), so $g$ is one degree “smoother” than $f$. Integration is a smoothing operation, a fact that reappears throughout analysis and signal processing.
For PhD-track students: The proof rests on continuity through the Squeeze Theorem and the Extreme Value Theorem, which is why the theorem as stated requires a continuous integrand. The Lebesgue version, where $g'(x) = f(x)$ holds only almost everywhere for an integrable $f$, and the precise role of absolute continuity, is the natural next chapter and reveals exactly how much of Part 1 survives beyond the Riemann setting.
Back to Integrals | Related: Riemann Sums | Next: FTC Part 2