1. The shock: two operations, one identity
If you've met the derivative and the integral as separate things, the relationship between them should feel implausible. Differentiation answers a local question: how steep is the curve right here? It looks at a single point and reports a rate. Integration answers a global question: what's the total area trapped between this curve and the axis, all the way across an interval? It looks at every point at once and reports an accumulation.
One is a microscope. The other is a tally. They are defined differently, they are computed differently, and they were studied separately for centuries — areas going back to Archimedes, tangent lines emerging again and again from optics, mechanics, and astronomy. There was no a-priori reason to suspect that finding the slope of one function would have anything at all to do with finding the area under another.
And yet:
Differentiation and integration are inverse operations. Each one undoes the other.
This is the Fundamental Theorem of Calculus. It comes in two parts — really two views of the same fact — and together they collapse two ancient problems into one machine.
Isaac Newton (working in the 1660s) and Gottfried Wilhelm Leibniz (working in the 1670s) discovered this connection independently, within a decade of each other, and from utterly different starting points — Newton from physics and fluxions, Leibniz from the symbolic manipulation of sums and differences. The bitter priority dispute that followed dominated English and Continental mathematics for a generation and arguably set British analysis back by a century. The dust settled long ago: both invented the calculus, both deserve the credit, and the notation we use today (dx, the elongated S of $\int$) is mostly Leibniz's. The theorem is the punchline they both arrived at.
2. Part 1: differentiation undoes integration
Let $f$ be continuous on $[a, b]$. Define the accumulation function
$$ F(x) = \int_a^x f(t)\,dt \qquad \text{for } x \in [a, b]. $$Then $F$ is differentiable on $(a, b)$ and
$$ F'(x) = f(x). $$Read it out loud. $F(x)$ is the area under $f$ from the fixed left endpoint $a$ up to a moving right endpoint $x$ — an area that grows as $x$ slides to the right. Part 1 says: if you differentiate that growing-area function, you get back the original $f$.
The dummy variable $t$ inside the integral is there because $x$ is already in use as the upper limit. The two roles must stay distinct. The variable being integrated over is $t$; the variable being held fixed (and later differentiated) is $x$. A common shorthand is to write $F(x) = \int_a^x f$, dropping the $\,dt$ once the convention is understood.
Two immediate consequences:
- Every continuous function has an antiderivative. The accumulation function $F$ is one — Part 1 hands it to you for free, whether or not you can write it down in elementary form.
- The integral with a variable upper limit is a function you can differentiate. So $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$, no matter what $a$ is.
If the upper limit is itself a function of $x$, you get a chain-rule version: $\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\,g'(x)$. The accumulation function is composed with $g$, and the derivative respects that composition.
3. Part 2: integration undoes differentiation
Let $f$ be continuous on $[a, b]$ and let $F$ be any antiderivative of $f$ on $[a, b]$ — that is, any function with $F'(x) = f(x)$. Then
$$ \int_a^b f(x)\,dx \;=\; F(b) - F(a). $$This is the workhorse. Part 1 is the theoretical statement that integration and differentiation are linked; Part 2 is what you actually use to compute integrals on paper. The recipe is just two lines:
- Find an antiderivative $F$ of $f$ — a function whose derivative is $f$.
- Evaluate $F(b) - F(a)$.
The standard notation for that subtraction is the bracket bar:
$$ \int_a^b f(x)\,dx = \Big[\,F(x)\,\Big]_a^b = F(b) - F(a). $$Note the strength of "any antiderivative." Antiderivatives are not unique — they form a family that differs only by an additive constant. But that constant cancels out when you subtract $F(b) - F(a)$, so the answer is the same no matter which member of the family you pick. We'll come back to this in the pitfalls.
Before Part 2, a definite integral was a limit of Riemann sums — a hard, slow, often impossible computation. After Part 2, it's a derivative problem run backward. Computing $\int_0^1 x^7\,dx$ becomes the trivial $\big[\tfrac{x^8}{8}\big]_0^1 = \tfrac{1}{8}$. The leap in computational power is enormous.
4. Why it works (geometric intuition)
The proof of Part 1 is almost a picture. Fix a positive continuous function $f$, and let $F(x)$ be the area under $f$ from $a$ to $x$. Ask: how fast is $F$ growing as $x$ moves to the right?
If $x$ advances by a tiny amount $dx$, the area picks up a thin sliver — a strip of width $dx$ whose height is approximately $f(x)$ (the function's value right there). The area of that strip is approximately
$$ dF \approx f(x)\,dx. $$Divide by $dx$:
$$ \frac{dF}{dx} \approx f(x). $$Take the limit as $dx \to 0$, where the approximation becomes exact (because $f$ is continuous, the strip's height converges to exactly $f(x)$), and you have Part 1. The rate at which area accumulates is the height of the curve. That's the whole secret.
For Part 2, the chain of reasoning is short. Part 1 says the accumulation function $F_a(x) = \int_a^x f$ is an antiderivative of $f$. Any other antiderivative $G$ differs from $F_a$ by a constant: $G(x) = F_a(x) + C$. Evaluate $G$ at the endpoints:
$$ G(b) - G(a) = \bigl(F_a(b) + C\bigr) - \bigl(F_a(a) + C\bigr) = F_a(b) - F_a(a) = \int_a^b f(x)\,dx - 0, $$since $F_a(a) = \int_a^a f = 0$. The constant cancels. So any antiderivative gives you the integral by simple subtraction. That is Part 2.
5. Using it in practice
Here is the entire workflow for a definite integral, distilled:
- Identify an antiderivative. Ask: what function, when differentiated, gives the integrand?
- Evaluate at the upper limit.
- Subtract the value at the lower limit.
Compute $\displaystyle \int_0^2 x^2\,dx$.
Step 1. An antiderivative of $x^2$ is $\tfrac{x^3}{3}$, because $\frac{d}{dx}\!\left[\tfrac{x^3}{3}\right] = x^2$.
Step 2. Set up the bracket evaluation:
$$ \int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2. $$Step 3. Subtract:
$$ \left[\frac{x^3}{3}\right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}. $$The area trapped between the parabola $y = x^2$ and the x-axis, from $x = 0$ to $x = 2$, is exactly $\tfrac{8}{3}$. No Riemann sums, no limits, no slicing. A single antiderivative and one subtraction.
Every technique you'll meet later — $u$-substitution, integration by parts, partial fractions, trig substitution — is at heart a way of finding the antiderivative. Once you have it, the Fundamental Theorem hands you the answer.
6. Common pitfalls
The constant of integration belongs in indefinite integrals — $\int x^2\,dx = \tfrac{x^3}{3} + C$ — because antiderivatives are determined only up to a constant. In a definite integral the constant cancels in the subtraction $F(b) - F(a)$, so writing $+C$ is harmless but pointless. Leave it off when evaluating endpoints; you'll get the same number either way.
By convention $\int_a^b f = -\int_b^a f$. If you mistakenly evaluate $F(a) - F(b)$ instead of $F(b) - F(a)$, you've not made an arithmetic error — you've computed the integral over the reversed interval. The magnitude is right; the sign is wrong. Always evaluate at the upper limit first.
The Fundamental Theorem applies when $f$ is continuous on the whole closed interval. If $f$ has a vertical asymptote inside $[a, b]$ — for example, $f(x) = 1/x^2$ on $[-1, 1]$ — naive application of Part 2 gives nonsense (a finite negative number for what is in fact a divergent improper integral). Discontinuities must be handled with improper integrals: split at the bad point, take limits, and check whether they converge.
In $F(x) = \int_a^x f(t)\,dt$, the variable $t$ is integrated over and disappears; the variable $x$ is what's left. Writing $\int_a^x f(x)\,dx$ uses $x$ for both roles and is a recipe for confusion. Pick different letters and keep them separate.
7. Worked examples
Try each one yourself before opening the solution. Each example exercises the same recipe — find an antiderivative, evaluate, subtract — under slightly different conditions.
Example 1 · Compute $\int_0^2 x^2\,dx$
Step 1. Antiderivative of $x^2$ is $\tfrac{x^3}{3}$.
Step 2. Evaluate at the endpoints:
$$ \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3}. $$Check. The parabola $y = x^2$ sweeps out an area that grows like a cube — and a cube of $2$, divided by three, is $\tfrac{8}{3}$.
Example 2 · Compute $\int_1^4 \frac{1}{x}\,dx$
Step 1. An antiderivative of $1/x$ is $\ln|x|$ (since $\tfrac{d}{dx}\ln x = \tfrac{1}{x}$ for $x > 0$). The interval $[1, 4]$ is in the positive region, so $\ln x$ suffices.
Step 2. Evaluate:
$$ \int_1^4 \frac{1}{x}\,dx = \big[\ln x\big]_1^4 = \ln 4 - \ln 1 = \ln 4 \approx 1.386. $$Note. The continuity condition matters: $1/x$ blows up at $x = 0$. As long as the interval avoids the origin we're safe.
Example 3 · Compute $\int_0^{\pi/2} \cos x\,dx$
Step 1. An antiderivative of $\cos x$ is $\sin x$.
Step 2. Evaluate:
$$ \int_0^{\pi/2} \cos x\,dx = \big[\sin x\big]_0^{\pi/2} = \sin\!\tfrac{\pi}{2} - \sin 0 = 1 - 0 = 1. $$Geometry. The area under one quarter-arch of cosine, from $0$ to $\pi/2$, is exactly $1$ — a clean result that would be unpleasant to derive from Riemann sums.
Example 4 · Use Part 1 to differentiate $G(x) = \int_2^x \sqrt{1 + t^4}\,dt$
This is a problem you cannot solve by finding an antiderivative — $\sqrt{1 + t^4}$ has no elementary antiderivative. But you don't need one. Part 1 says: the derivative of an accumulation function is the integrand evaluated at the upper limit.
Answer.
$$ G'(x) = \sqrt{1 + x^4}. $$That's all there is to it. Part 1 turns a hard-looking question into a one-step substitution.
Example 5 · Compute $\int_{-1}^{1} (3x^2 + 2x)\,dx$
Step 1. Use linearity. An antiderivative of $3x^2 + 2x$ is $x^3 + x^2$ (since $\frac{d}{dx}(x^3 + x^2) = 3x^2 + 2x$).
Step 2. Evaluate:
$$ \big[x^3 + x^2\big]_{-1}^{1} = (1 + 1) - ((-1) + 1) = 2 - 0 = 2. $$Sanity check. The integrand $3x^2$ is even (contributes positively across the symmetric interval) and $2x$ is odd (contributes zero by symmetry). So the answer should be $\int_{-1}^{1} 3x^2\,dx = 2 \cdot \int_0^1 3x^2\,dx = 2 \cdot 1 = 2$. ✓