Fundamental Theorem of Calculus

Calculus is the math of change. One half measures how fast something is changing at a single instant, and the other half adds up small amounts to give a total.

Fundamental Theorem of Calculus: the central idea

The Fundamental Theorem connects accumulation and instantaneous rate: differentiation can recover a continuous integrand, and antiderivatives can evaluate accumulated change.

Words you need

Definite integral

A definite integral is a single number that measures the signed area between a curve and the horizontal axis across a fixed interval from $a$ to $b$.

Antiderivative

An antiderivative $F$ of a function $f$ is a function whose derivative is $f$, and the theorem says any one of them will do because their differences cancel.

Accumulation function

An accumulation function is a running total written as $A(x)=\int_a^x f(t)\,dt$, and Part 1 says its derivative is simply $f(x)$.

Net change

Net change is the total effect of a rate over an interval, so integrating a speed gives displacement and integrating a flow rate gives the volume added.

Signed area

Signed area counts the region above the axis as positive and the region below as negative, which is why an integral can come out to zero even when the curve is far from flat.

Evaluation bar

The evaluation bar, written $[F(x)]_a^b$, is just shorthand for the instruction to compute $F(b)-F(a)$.

What to know before this lesson

Understand limits, continuity, antiderivatives, definite integrals, and signed area.

If one of those prerequisites is uncertain, use the Calculus subject guide to locate the earlier concept before memorizing a procedure.

The fundamental theorem of calculus: a worked example

Every step, with the arithmetic

1. Step 1 - Write the definite integral$\int_1^3 (3x^2+2)\,dx$
2. Step 2 - Find an antiderivative$3x^2$ comes from $x^3$, and $2$ comes from $2x$, so $F(x)=x^3+2x$
3. Step 3 - Confirm $F$ is right$F'(x)=3x^2+2$, which matches the integrand exactly
4. Step 4 - Plug in the top limit$F(3)=3^3+2(3)=27+6=33$
5. Step 5 - Plug in the bottom limit$F(1)=1^3+2(1)=1+2=3$
6. Step 6 - Subtract, top minus bottom$33-3=30$
7. Step 7 - See why $+C$ never matters$(33+C)-(3+C)=30$, so the constant always cancels in a definite integral
8. Step 8 - Rough check on the sizeThe curve rises from height $5$ at $x=1$ to height $29$ at $x=3$; a strip of width $2$ and average height about $15$ is roughly $30$

Because an antiderivative of $3x^2$ is $x^3$, $\int_0^2 3x^2dx=2^3-0^3=8$.

Definite integrals worked all the way to a number

The step-by-step method for the fundamental theorem of calculus

1. For a definite integral, find an antiderivative $F$ of the integrand.
2. Evaluate $F(b)-F(a)$ in upper-minus-lower order and retain exact values when possible.
3. Interpret the signed accumulation in context; split the interval if geometric area rather than net change is requested.

How to check your answer

Differentiate $x^3$ to recover $3x^2$, and estimate that the positive area on an interval of width $2$ should be positive and finite.

Compare units and qualitative behavior. A derivative should match the graph's slope, while a definite integral should agree with the signed area and scale of the interval.

A mistake that changes the mathematics

A definite integral is a number and does not receive $+C$. The arbitrary constant cancels in $F(b)-F(a)$.

Where you will actually use this

Odometer versus speedometer

The speedometer shows a rate and the odometer shows the accumulated total. Integrating the speed over a trip gives the distance, and differentiating the odometer reading gives back the speed.

Water in a reservoir

Sensors record inflow in litres per minute. Integrating that rate across a storm gives the exact number of litres added without measuring the tank.

Electric charge from current

Current is the rate charge flows, so engineers integrate a current curve over time to find the total charge a battery delivered.

How the fundamental theorem of calculus connects to the rest of calculus

• Integration techniques — This theorem is useless without an antiderivative, and those techniques are how you find one.
• Derivative rules — Part 1 pairs with the chain rule whenever the integral's upper limit is a function of $x$ rather than plain $x$.
• Understanding limits — A definite integral is defined as the limit of a sum of thin rectangles, so limits are what make the area exact instead of an estimate.

Try a transfer problem

Differentiate $G(x)=\int_1^{x^2}\cos t\,dt$ and explain why the chain rule contributes a factor of $2x$.

Show the worked answer

Write $G(x)=\int_1^{x^2}\cos t\,dt$ as a composition. Let $H(u)=\int_1^{u}\cos t\,dt$, so $G(x)=H(x^2)$. Part 1 of the Fundamental Theorem says $H'(u)=\cos u$, because differentiating a running total just hands back the height of the curve at the top edge. Now the chain rule finishes the job: $G'(x)=H'(x^2)\cdot\frac{d}{dx}(x^2)=\cos(x^2)\cdot 2x=2x\cos(x^2)$. The factor of $2x$ appears because the upper limit is not creeping along at speed $1$. When $x$ grows by a little, the top edge $x^2$ moves $2x$ times as fast, so the accumulated area grows $2x$ times faster than it would for a plain upper limit of $x$. You can confirm it the long way: an antiderivative of $\cos t$ is $\sin t$, so $G(x)=\sin(x^2)-\sin(1)$, and differentiating that gives $2x\cos(x^2)$ as well.

Work without copying the example. When finished, use the relevant focused calculator or formula reference to check the setup and result, then correct the first line where your reasoning changed. When the method feels reliable, move to calculus practice questions.

Questions about the fundamental theorem of calculus

What is the difference between Part 1 and Part 2?

Part 1 starts with an integral and differentiates it: $\frac{d}{dx}\int_a^x f(t)\,dt=f(x)$. Part 2 starts with a derivative and integrates it: $\int_a^b f(x)\,dx=F(b)-F(a)$. Part 1 is the theory that proves antiderivatives exist; Part 2 is the calculating tool.

Why can I pick any antiderivative I like?

Two antiderivatives of the same function differ only by a constant. When you subtract $F(b)-F(a)$ that constant appears in both terms and cancels, so $x^3$ and $x^3+100$ give the same answer for $\int_0^2 3x^2 dx$, namely $8$.

If the integral is zero, does that mean the area is zero?

No. It means the area above the axis exactly matches the area below it. $\int_{-1}^{1}x^3dx=0$, yet the graph clearly encloses real regions. For true geometric area, split the interval at each place the curve crosses the axis and add the absolute values.

Does the theorem still work if the function has a jump in the interval?

Not as stated. Part 2 requires the integrand to be continuous on the closed interval $[a,b]$. Blindly applying it to $\int_{-1}^{1}\frac{1}{x^2}dx$ gives $-2$, which is impossible for a curve that never dips below the axis. Break the interval at the trouble spot and handle each piece as an improper integral instead.