# Fundamental Theorem of Calculus Part 1: Derivative of an Integral

> FTC part 1: if f is continuous, the derivative of the integral from a to x of f(t) dt is f(x). Statement, graph, chain rule version and worked examples.

Source: https://practicecalculus.com/theorems/fundamental-theorem-of-calculus-part-1
Updated: 2026-11-03

$$ \frac{d}{dx} \int_a^x f(t)\,dt = f(x) $$

## At a glance

- **Hypotheses:** $f$ is continuous on $[a, b]$, and $F(x) = \int_a^x f(t)\,dt$ for $x$ in $[a, b]$
- **Conclusion:** $F$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $F'(x) = f(x)$
- **In words:** The rate at which accumulated area grows is the height of the curve
- **Chain rule version:** $\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\,g'(x)$
- **Big consequence:** Every continuous function has an antiderivative
- **Course:** Calculus 1

## Key takeaways

- Differentiating an integral with a variable upper limit gives back the integrand evaluated at that limit.
- When the upper limit is a function $g(x)$, multiply by $g'(x)$: this is the chain rule.
- The lower limit $a$ does not affect the derivative, as long as it is a constant.
- FTC 1 shows that differentiation and integration are inverse processes.

The **fundamental theorem of calculus, part 1** (FTC 1) says that if you build a function by accumulating area under a continuous curve, the derivative of that function is the curve itself. It is the first half of the link between derivatives and integrals.

## Statement of FTC part 1

Let $f$ be **continuous** on $[a, b]$, and define the **accumulation function**

$$F(x) = \int_a^x f(t)\,dt, \qquad a \le x \le b$$

Then $F$ is continuous on $[a, b]$, differentiable on $(a, b)$, and

$$F'(x) = f(x)$$

In Leibniz notation, $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$. The variable $t$ is a dummy variable of integration, and $x$ is the upper limit.

## What FTC 1 means

$F(x)$ is the signed area under $y = f(t)$ from $a$ to $x$. Nudge $x$ a little to the right by $h$, and the new area is a thin strip of width $h$ and height about $f(x)$. So

$$\frac{F(x + h) - F(x)}{h} \approx \frac{f(x)\,h}{h} = f(x)$$

and the approximation becomes exact as $h \to 0$. The rate at which area accumulates equals the height of the curve.

```graph
title: 'FTC part 1: area under y = x² from 0 to x grows like F(x) = x³/3'
caption: 'The shaded area up to $x = 1.5$ is $F(1.5) = \frac{1.5^3}{3} = 1.125$. The dashed curve $F(x) = \frac{x^3}{3}$ tracks that area, and its slope at each $x$ equals the height $x^2$.'
x: [0, 2.2]
y: [0, 4.5]
functions:
  - expr: x^2
    label: y = f(x) = x²
  - expr: x^3/3
    label: F(x) = x³/3 (accumulated area)
    dashed: true
shade: {expr: x^2, from: 0, to: 1.5}
points:
  - [1.5, 1.125, 'F(1.5) = 1.125']
```

## Why continuity matters

Let $f(t) = 0$ for $t < 0$ and $f(t) = 1$ for $t \ge 0$, and let $F(x) = \int_{-1}^x f(t)\,dt$. The accumulated area is

$$F(x) = \begin{cases} 0 & x < 0 \\ x & x \ge 0 \end{cases}$$

which has a corner at $x = 0$. So $F'(0)$ does not exist, and certainly does not equal $f(0) = 1$. The integral still makes sense, but at the jump the conclusion $F' = f$ fails. Away from the jump, where $f$ is continuous, $F' = f$ still holds.

## Worked examples

**Example 1: the basic form.**

$$\frac{d}{dx}\int_1^x \sqrt{1 + t^3}\,dt = \sqrt{1 + x^3}$$

No integration needed, and none possible in elementary terms. Replace $t$ with $x$ and you are done.

**Example 2: upper limit is a function of $x$ (chain rule).** Find $\frac{d}{dx}\int_0^{x^2}\cos t\,dt$.

**Step 1.** Let $u = x^2$. Then the integral is $G(u) = \int_0^u \cos t\,dt$, and by FTC 1, $G'(u) = \cos u$.

**Step 2.** By the [chain rule](/blog/chain-rule), $\frac{d}{dx}G(x^2) = \cos(x^2)\cdot 2x$.

$$\frac{d}{dx}\int_0^{x^2}\cos t\,dt = 2x\cos(x^2)$$

Check: the integral equals $\sin(x^2)$, whose derivative is $2x\cos(x^2)$.

**Example 3: both limits depend on $x$.** Find $\frac{d}{dx}\int_x^{2x} e^{t^2}\,dt$.

**Step 1.** Split at a constant: $\int_x^{2x} = \int_0^{2x} - \int_0^{x}$.

**Step 2.** Differentiate each piece with FTC 1 and the chain rule:

$$\frac{d}{dx}\int_x^{2x} e^{t^2}\,dt = e^{(2x)^2}\cdot 2 - e^{x^2}\cdot 1 = 2e^{4x^2} - e^{x^2}$$

**Example 4: a lower limit with $x$.** Flip the limits first: $\frac{d}{dx}\int_x^5 \sin(t^2)\,dt = \frac{d}{dx}\left(-\int_5^x \sin(t^2)\,dt\right) = -\sin(x^2)$.

## Common mistakes

1. **Forgetting the chain rule factor.** $\frac{d}{dx}\int_0^{x^2}\cos t\,dt$ is $2x\cos(x^2)$, not $\cos(x^2)$.
2. **Plugging in the lower limit too.** For $\int_a^x f(t)\,dt$ with constant $a$, the answer is just $f(x)$. The constant lower limit only shifts $F$ by a constant.
3. **Forgetting the sign when $x$ is in the lower limit.** Swapping the limits introduces a minus sign.
4. **Integrating first when it is impossible.** Integrands like $e^{t^2}$ have no elementary antiderivative, but FTC 1 still differentiates the integral instantly.

## Why FTC 1 matters

FTC 1 proves that every continuous function has an antiderivative, namely $\int_a^x f(t)\,dt$. That fact is what makes [FTC part 2](/theorems/fundamental-theorem-of-calculus-part-2) work, turning definite integrals into simple subtraction. It also defines important functions such as $\ln x = \int_1^x \frac{1}{t}\,dt$. For the full story of both parts, read the [fundamental theorem of calculus guide](/blog/fundamental-theorem-of-calculus), and see [Riemann sums](/blog/riemann-sums) for the definition of the integral it relies on.

## Frequently asked questions

### What does the first fundamental theorem of calculus say?

If $f$ is continuous and $F(x) = \int_a^x f(t)\,dt$, then $F'(x) = f(x)$. The derivative of the accumulated area function is the original function.

### How do you take the derivative of an integral with x in the upper limit?

Replace $t$ in the integrand by the upper limit. If the upper limit is $g(x)$ instead of $x$, also multiply by $g'(x)$. For example $\frac{d}{dx}\int_0^{x^2}\cos t\,dt = 2x\cos(x^2)$.

### What if both limits of integration contain x?

Split the integral at a constant, or use the formula $\frac{d}{dx}\int_{h(x)}^{g(x)} f(t)\,dt = f(g(x))\,g'(x) - f(h(x))\,h'(x)$.

### What is the difference between FTC part 1 and part 2?

Part 1 differentiates an integral: $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$. Part 2 evaluates a definite integral with an antiderivative: $\int_a^b f(x)\,dx = F(b) - F(a)$.