The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus — Moosa Academy

Finding the area under a curved graph looks like a hard geometry problem — there is no formula for a shape with a curved top. The Fundamental Theorem removes the geometry entirely: find one function, substitute two numbers, subtract. That is the whole calculation.

Theorem The statement

The shaded region sits under

f(x)

, between

x = a

and

x = b

. Its area is given by a single subtraction.

\int_{a}^{b} f(x)\,dx = F(b) - F(a)

where

F(x)

is any integral of

f(x)

Four steps every time: integrate $f$ to get $F$ , substitute $b$ , substitute $a$ , subtract.

Concept Why this is powerful

Measuring a curved region directly is genuinely difficult — the usual area formulas all assume straight edges. The theorem trades that problem for a different one:

Geometry — area under an arbitrary curve: very hard

Algebra — substitute two numbers and subtract: easy

However complicated the curve, once $F$ is known the area between any two points is only substitution. That exchange — geometry for algebra — is what makes the theorem fundamental.

Example A straight line: f(x) = x from 3 to 7

Step 1 — integrate:

F(x) = \dfrac{x^{2}}{2}

Step 2 — substitute

b = 7

F(7) = \dfrac{49}{2}

Step 3 — substitute

a = 3

F(3) = \dfrac{9}{2}

Step 4 — subtract:

\dfrac{49}{2} - \dfrac{9}{2} = \dfrac{40}{2} = 20

⟹ area = 20

This one can be checked by hand: the region is a trapezium with parallel sides 3 and 7 and width 4, giving $\tfrac{1}{2}(3+7)(4) = 20$ . The theorem agrees — which is reassuring before trusting it on shapes that cannot be checked so easily.

Example A curve: f(x) = x³ from 1 to 3

Step 1 — integrate:

F(x) = \dfrac{x^{4}}{4}

Step 2 — substitute

b = 3

F(3) = \dfrac{81}{4}

Step 3 — substitute

a = 1

F(1) = \dfrac{1}{4}

Step 4 — subtract:

\dfrac{81}{4} - \dfrac{1}{4} = \dfrac{80}{4} = 20

⟹ area = 20

Here there is no elementary shape to fall back on — the top edge is a cubic curve. Yet the work was no harder than the first example. The two answers happening to agree at 20 is a coincidence, but a useful one: the same four steps handled a straight line and a curve identically.

Note Where the constant C went

Indefinite integrals always carry $+ C$ , yet neither example above used one. Keeping it shows why:

\bigl(F(b) + C\bigr) - \bigl(F(a) + C\bigr) = F(b) - F(a)

The same constant appears in both terms and cancels in the subtraction. It makes no difference which member of the family is chosen, so for a definite integral $C$ is simply left out.

Summary

The area under f(x) between a and b equals F(b) − F(a).
Four steps: integrate, substitute b, substitute a, subtract.
For f(x) = x from 3 to 7: 49/2 − 9/2 = 20.
For f(x) = x³ from 1 to 3: 81/4 − 1/4 = 20.
The constant C cancels in the subtraction, so definite integrals omit it.
A hard geometry problem becomes ordinary algebra — that is why the theorem is fundamental.