What Is The Fundamental Theorem of Calculus?
The Fundamental Theorem of Calculus is the result that ties differentiation and integration together as inverse operations. It comes in two parts. Part 1 tells you how to differentiate an integral; Part 2 tells you how to evaluate one.
Part 1 (the derivative form). If $f$ is continuous on $[a, b]$ and we define the accumulation function
$$g(x) = \int_a^x f(t),dt,$$
then $g$ is differentiable on $(a, b)$ and
$$g'(x) = f(x).$$
In words, the rate at which accumulated area grows at $x$ is exactly the height of the curve at $x$.
Part 2 (the evaluation form). If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ (a function with $F' = f$), then
$$\int_a^b f(x),dx = F(b) - F(a).$$
This is the workhorse of every calculus course: to find a definite integral, find an antiderivative and subtract its values at the two ends. The two parts together are why calculus is one subject and not two. For the wider map of the field, see the calculus overview.
What Does Part 1 Of The Fundamental Theorem of Calculus Say?
Part 1 is about the accumulation function $g(x) = \int_a^x f(t),dt$. Read it as "the signed area under $f$ from the fixed start $a$ up to the moving point $x$." As $x$ slides right, $g$ grows by adding thin slivers of area, and the height of each new sliver is just $f(x)$.
That is the whole claim: the derivative of the area-so-far is the height of the curve.
$$\frac{d}{dx}\int_a^x f(t),dt = f(x)$$
The letter $t$ inside the integral is a dummy variable. It is only there to sweep from $a$ to $x$, and it never survives into the answer. The output depends on the upper limit $x$, not on $t$.
Often the upper limit is itself a function $u(x)$ rather than a bare $x$. Then Part 1 combines with the chain rule:
$$\frac{d}{dx}\int_a^{u(x)} f(t),dt = f\big(u(x)\big)\cdot u'(x)$$
You evaluate the integrand at the top limit, then multiply by the derivative of that top limit. Missing the $u'(x)$ factor is the single most common slip on this topic, and it has its own entry in the mistakes section below. When the moving limit is the lower one instead, a sign appears: $\dfrac{d}{dx}\int_{u(x)}^{b} f(t),dt = -,f\big(u(x)\big)\cdot u'(x)$.
What Does Part 2 Of The Fundamental Theorem of Calculus Say?
Part 2, also called the evaluation theorem or the Newton-Leibniz formula, is the part you use to get a number out of a definite integral.
$$\int_a^b f(x),dx = F(b) - F(a), \qquad \text{where } F'(x) = f(x)$$
A common shorthand writes $F(b) - F(a)$ as $\big[F(x)\big]_a^b$. The continuity of $f$ on the closed interval $[a, b]$ is what guarantees an antiderivative exists and that this subtraction is valid. Skip that check and the formula can hand you an absurd answer, as the mistakes section shows.
Two consequences are worth naming early:
The constant does not matter. Any antiderivative works, because if $F$ and $G$ both differentiate to $f$, they differ by a constant that cancels in $F(b) - F(a)$. This is why there is no $+C$ on a definite integral.
Swapping the limits flips the sign: $\int_a^b f = -\int_b^a f$. Reversing the direction of travel negates the accumulation.
For the mechanics of finding the antiderivative $F$ in the first place, the integration reference collects the standard rules.
How Do You Prove The Fundamental Theorem of Calculus?
Both parts follow from one idea: a thin slice of accumulated area is almost a rectangle whose height is the value of the function.
Proof sketch of Part 1. Start from the limit definition of the derivative of $g(x) = \int_a^x f(t),dt$:
$$g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} = \lim_{h \to 0} \frac{1}{h}\int_x^{x+h} f(t),dt$$
The integral $\int_x^{x+h} f(t),dt$ is the area of a sliver of width $h$. The Mean Value Theorem for integrals says that because $f$ is continuous, there is some point $c$ between $x$ and $x+h$ where that sliver equals a clean rectangle: $\int_x^{x+h} f(t),dt = f(c)\cdot h$. Dividing by $h$ leaves $f(c)$.
$$g'(x) = \lim_{h \to 0} f(c), \qquad x \le c \le x+h$$
As $h \to 0$, the point $c$ is squeezed toward $x$, and because $f$ is continuous, $f(c) \to f(x)$. Therefore $g'(x) = f(x)$, which is Part 1.
Proof sketch of Part 2. Let $F$ be any antiderivative of $f$, and let $g(x) = \int_a^x f(t),dt$ be the accumulation function from Part 1, so $g' = f$. Two functions with the same derivative differ by a constant, so $F(x) = g(x) + C$. Now evaluate the difference at the endpoints:
$$F(b) - F(a) = \big[g(b) + C\big] - \big[g(a) + C\big] = g(b) - g(a)$$
But $g(a) = \int_a^a f(t),dt = 0$ and $g(b) = \int_a^b f(t),dt$, so $F(b) - F(a) = \int_a^b f(x),dx$. That is Part 2, obtained directly from Part 1.
How Do You Use The Fundamental Theorem of Calculus? Worked Examples
Each example below is fully stepped, and every definite integral is checked by differentiating the antiderivative back to the integrand.
Example 1: Differentiate an accumulation function (Part 1).
Find $g'(x)$ for $g(x) = \int_0^x \left(3t^2 + 1\right),dt$.
By Part 1, differentiating the accumulation function just returns the integrand evaluated at the upper limit:
$$g'(x) = 3x^2 + 1$$
Check: compute $g$ directly, $g(x) = \left[t^3 + t\right]_0^x = x^3 + x$, and differentiate: $g'(x) = 3x^2 + 1$. The two agree.
Final answer: $g'(x) = 3x^2 + 1$.
Example 2: A variable upper limit with the chain rule (Part 1).
Find $\dfrac{d}{dx}\displaystyle\int_1^{x^2} \ln t,dt$.
The upper limit is $u(x) = x^2$, so evaluate the integrand at $u(x)$ and multiply by $u'(x) = 2x$:
$$\frac{d}{dx}\int_1^{x^2} \ln t,dt = \ln\big(x^2\big)\cdot 2x = 2x\ln\big(x^2\big)$$
Final answer: $2x\ln\big(x^2\big)$ (equivalently $4x\ln x$ for $x > 0$).
Example 3: Evaluate a definite integral (Part 2).
Evaluate $\displaystyle\int_0^{\pi} \sin x,dx$.
An antiderivative of $\sin x$ is $F(x) = -\cos x$. Apply Part 2:
$$\int_0^{\pi} \sin x,dx = \big[-\cos x\big]_0^{\pi} = \big(-\cos \pi\big) - \big(-\cos 0\big) = \big(1\big) - \big(-1\big) = 2$$
Check: $F'(x) = \dfrac{d}{dx}\left(-\cos x\right) = \sin x$, the original integrand, so the antiderivative is correct.
Final answer: $\displaystyle\int_0^{\pi} \sin x,dx = 2$.
Example 4: A polynomial definite integral (Part 2).
Evaluate $\displaystyle\int_1^{3} 2x,dx$.
An antiderivative of $2x$ is $F(x) = x^2$. Apply Part 2:
$$\int_1^{3} 2x,dx = \big[x^2\big]_1^{3} = 3^2 - 1^2 = 9 - 1 = 8$$
Check: $\dfrac{d}{dx}\left(x^2\right) = 2x$, the integrand, so the evaluation is valid.
Final answer: $\displaystyle\int_1^{3} 2x,dx = 8$.
What Are The Standard Antiderivatives For The Fundamental Theorem of Calculus?
Part 2 is only as easy as your ability to spot an antiderivative $F$. These are the ones that cover most first-year definite integrals. Each is stated as an indefinite integral, so it carries $+C$; that constant cancels the moment you evaluate between two limits.
Table: Standard antiderivatives used with Part 2 of the Fundamental Theorem of Calculus.
Function $f(x)$ | Antiderivative $F(x)$ | Note |
|---|---|---|
$x^n$ (with $n \neq -1$) | $\dfrac{x^{n+1}}{n+1} + C$ | the power rule for integration |
$\dfrac{1}{x}$ | $\ln\lvert x\rvert + C$ | the $n = -1$ exception |
$e^x$ | $e^x + C$ | its own antiderivative |
$\sin x$ | $-\cos x + C$ | sign flips |
$\cos x$ | $\sin x + C$ | |
$\sec^2 x$ | $\tan x + C$ | reverse of the tangent derivative |
When the integrand is a product that does not match a single row, a method such as integration by parts rewrites it into pieces the table can handle.
Why Does The Fundamental Theorem of Calculus Work?
The theorem feels surprising the first time, because differentiation (slopes) and integration (areas) look unrelated. The bridge is the accumulation function, and the reason it works is geometric.
Area grows at the speed of the height. If the curve is tall at $x$, each new vertical strip of area added as $x$ moves right is tall, so the accumulated area is increasing fast. If the curve is short, the area barely grows. The rate of area growth is the height, which is exactly $g'(x) = f(x)$.
A total is the sum of its rates. Distance is the running total of speed; charge is the running total of current; volume is the running total of flow. Part 2 says the net total between $a$ and $b$ is just the change in the quantity whose rate you integrated, $F(b) - F(a)$.
Continuity keeps the strips honest. The proof needed a point $c$ where a thin sliver equalled a clean rectangle. That is guaranteed only when $f$ has no jumps or breaks on the interval, which is why continuity is a hypothesis and not a technicality.
Seen this way, "the derivative of the integral" and "the integral of the derivative" are the same statement read forwards and backwards. One direction builds a total from rates; the other recovers the rate from the total.
Who Discovered The Fundamental Theorem of Calculus?
The link between areas and slopes was glimpsed by several people across the 1600s before two rivals gave it a full theory on their own.
Two earlier figures set the stage:
James Gregory (1638–1675, Scotland) published a first rudimentary version of the theorem, recognising the inverse link between tangents and areas.
Isaac Barrow (1630–1677, England), Newton's own teacher at Cambridge, proved a geometric form of the result before handing the chair of mathematics to his student.
Later, Augustin-Louis Cauchy (1789–1857, France) and Bernhard Riemann gave the integral its rigorous modern definition, so the theorem finally rested on a precise notion of area rather than intuition.
Where Is The Fundamental Theorem of Calculus Used In The Real World?
The theorem is the reason "total from a rate" is a one-line calculation across the sciences.
Physics and motion: the distance a body travels is the integral of its speed, and its speed is the derivative of distance. Every kinematics problem is Part 1 and Part 2 in disguise.
Engineering and flow: the volume of water through a pipe, or charge through a wire, is the integral of the flow rate or current over time, evaluated by $F(b) - F(a)$.
Economics: total cost is the integral of marginal cost, and total revenue the integral of marginal revenue, so a firm recovers totals from the per-unit rates it can measure.
Probability and statistics: the probability that a continuous variable lands in an interval is the integral of its density, and the cumulative distribution function is the accumulation function of Part 1.
Signal processing and graphics: smoothing, filtering, and shading all rest on integrating a rate and differentiating an accumulated signal, the two moves the theorem makes interchangeable.
One theorem lets every field swap freely between a quantity and its rate of change, which is why it earns the word "fundamental."
What Are The Most Common Mistakes With The Fundamental Theorem of Calculus?
These four errors account for most lost marks on the theorem, and each matches a question real students ask on r/calculus, r/learnmath, and course common-error handouts.
Forgetting the inner derivative in Part 1.
Where it slips in:
When the upper limit is a function such as $x^2$, a student writes $\dfrac{d}{dx}\int_1^{x^2}\ln t,dt = \ln(x^2)$ and stops.
Don't do this:
Do not treat a variable upper limit as if it were a bare $x$. The chain rule still applies.
The correct way:
Evaluate the integrand at the top limit, then multiply by the derivative of that limit: $\ln(x^2)\cdot 2x = 2x\ln(x^2)$.
Swapping the bounds without flipping the sign.
Where it slips in:
A student rewrites $\int_a^b$ as $\int_b^a$ to make the numbers friendlier, or differentiates an integral with the variable on the lower limit and keeps it positive.
Don't do this:
Do not reorder the limits for free. Reversing them negates the integral: $\int_a^b f = -\int_b^a f$.
The correct way:
Keep the order, or flip the sign when you flip the limits. For a variable lower limit, $\dfrac{d}{dx}\int_{x}^{5} f(t),dt = -f(x)$.
Applying Part 2 across a discontinuity.
Where it slips in:
A student evaluates $\int_{-1}^{1}\dfrac{1}{x^2},dx$ as $\left[-\dfrac{1}{x}\right]_{-1}^{1} = -1 - 1 = -2$.
Don't do this:
Do not apply Part 2 when $f$ is not continuous on the whole interval. Here $\dfrac{1}{x^2}$ blows up at $x = 0$.
The correct way:
Check continuity first. Because the integrand is positive everywhere, a negative answer is impossible; the integral is improper and in fact diverges, so no finite value from Part 2 is valid.
Letting the dummy variable clash with the limit.
Where it slips in:
A student writes the accumulation function as $\int_a^x f(x),dx$, using $x$ for both the moving limit and the variable of integration.
Don't do this:
Do not reuse the limit's letter inside the integral. It hides which $x$ is which and breaks Part 1.
The correct way:
Use a fresh dummy letter: $g(x) = \int_a^x f(t),dt$. The $t$ sweeps the interval; the $x$ is the output variable.
Practice Problems On The Fundamental Theorem of Calculus
Work each one, then check against the answer. Answers are verified.
Find $g'(x)$ for $g(x) = \int_0^x \left(t^2 + 2t\right),dt$.
(Answer: $g'(x) = x^2 + 2x$.)Find $\dfrac{d}{dx}\displaystyle\int_2^{x^3}\sin t,dt$.
(Answer: $\sin(x^3)\cdot 3x^2 = 3x^2\sin(x^3)$.)Evaluate $\displaystyle\int_0^{2} 3x^2,dx$.
(Answer: $\left[x^3\right]_0^2 = 8$.)Evaluate $\displaystyle\int_1^{4}\dfrac{1}{\sqrt{x}},dx$.
(Answer: $\left[2\sqrt{x}\right]_1^4 = 4 - 2 = 2$.)Evaluate $\displaystyle\int_0^{\pi/2}\cos x,dx$.
(Answer: $\left[\sin x\right]_0^{\pi/2} = 1$.)Find $\dfrac{d}{dx}\displaystyle\int_{x}^{5} e^t,dt$.
(Answer: the moving limit is the lower one, so $-e^x$.)
Where Should You Go Next After The Fundamental Theorem of Calculus?
The theorem is the hinge of calculus, and several natural doors open from here.
Integration. Build the antiderivative skills Part 2 depends on, from the power rule to substitution.
The derivative. Strengthen the differentiation side that Part 1 reverses, including the chain rule you need for variable limits.
Sequences. The Riemann sums behind the integral are limits of sequences, so this sharpens the idea of area as a limit.
If your child is meeting the Fundamental Theorem of Calculus for the first time, a live Bhanzu trainer teaches it from the accumulation picture up, so the two parts feel like one idea, in the Bhanzu math program.
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