The Fundamental Theorem of Calculus

Shade the area under a curve from a fixed start a to a moving end x, and that area becomes a function of x. The page draws it below the curve, measures how fast it grows, and puts that rate beside the height of the curve at x. Part 1 of the theorem says they are always the same number, and part 2 follows from it: an antiderivative evaluates a definite integral.

Part 1	$\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$
Part 2	$\int_{a}^{b} f (x) d x = F (b) - F (a), F^{'} = f$
With x² or any g(x) at the top	$\frac{d}{d x} \int_{a}^{g (x)} f (t) d t = f (g (x)) g^{'} (x)$

Worked example — The Fundamental Theorem of Calculus

Area from a to x, by Simpson's rule	0.7468
Height of the curve at x	0.3679

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Lesson

The theory — The Fundamental Theorem of Calculus

The accumulation function of f from a is A(x) = ∫ₐˣ f(t) dt, the signed area under f between a and x. The fundamental theorem of calculus has two parts: A′(x) = f(x) wherever f is continuous, and therefore ∫ₐᵇ f(x) dx = F(b) − F(a) for any antiderivative F of f.

What each symbol means

A(x): the area under f from a to x, its sign flips below the axis, and flips again when x is left of a
a: the fixed start; changing it moves A by a constant and leaves A′ alone

Where the formula comes from

Part 1: the slope of the area is the height of the curve. Over a short step from x to x + h the extra area is a strip of width h whose height is close to f(x), so the area grows at f(x) per unit of x.
$\frac{A ( x + h ) - A ( x )}{h} = \frac{1}{h} \int_{x}^{x + h} f (t) d t ⟶ f (x)$
Part 2: evaluate with any antiderivative. A and any antiderivative F have the same derivative, so they differ by a constant, and A(a) = 0 fixes it.
$A (x) = F (x) + C, A (a) = 0 \Rightarrow C = - F (a) \Rightarrow \int_{a}^{b} f = F (b) - F (a)$

Assumes: f is continuous on the interval. Where f jumps, the area function has a corner and no slope at that point, though it is still defined.

Problems solved in full

Differentiate an integral with x at the top 1 steps
Find d/dx ∫₁^x √(1 + t³) dt.
1. Part 1, as it stands: the derivative of the area function is the integrand at x.
  
  $\frac{d}{d x} \int_{1}^{x} 1 + t^{3} d t = 1 + x^{3}$
Answer

√(1 + x³). No antiderivative of √(1 + t³) is needed, which is lucky: it has no elementary one.

$1 + x^{3}$
x² at the top: the chain rule joins in 2 steps
Find d/dx ∫₀^x² cos t dt.
Show solution
1. The area function is A(u) with u = x², so differentiate A at u and multiply by du/dx:
  
  $\frac{d}{d x} \int_{0}^{x^{2}} cos t d t = cos (x^{2}) \cdot 2 x$
2. Check by doing the integral first:
  
  $\int_{0}^{x^{2}} cos t d t = sin (x^{2}) \Rightarrow \frac{d}{d x} sin (x^{2}) = 2 x cos (x^{2})$
Answer

2x cos(x²). Writing cos(x²) alone forgets the inner derivative, exactly as it would in any chain rule.

$2 x cos (x^{2})$
x in both limits 3 steps
Find d/dx ∫_x^2x t² dt.
Show solution
1. Split at any fixed point, say 0, so each piece has x in one limit only:
  
  $\int_{x}^{2 x} t^{2} d t = \int_{0}^{2 x} t^{2} d t - \int_{0}^{x} t^{2} d t$
2. Differentiate each piece, with the chain rule on the upper one:
  
  $\frac{d}{d x} = (2 x)^{2} \cdot 2 - x^{2} = 8 x^{2} - x^{2} = 7 x^{2}$
3. Check by doing the integral first:
  
  $\int_{x}^{2 x} t^{2} d t = \frac{8 x ^{3} - x ^{3}}{3} = \frac{7 x ^{3}}{3} \Rightarrow 7 x^{2}$
Answer

7x². The lower limit enters with a minus sign, because increasing it removes area.

$7 x^{2}$
A definite integral from an antiderivative 1 steps
Evaluate ∫₀^π sin x dx.

Open this problem in the tool
Show solution
1. −cos x is an antiderivative of sin x, so part 2 applies:
  
  $\int_{0}^{π} sin x d x = [- cos x]_{0}^{π} = - cos π + cos 0 = 1 + 1$
Answer

2. Open the problem in the tool: the area found slice by slice is 2 as well, and the area function is flat at π, its highest point.

$\int_{0}^{π} sin x d x = 2$
Reading an area function from its integrand 3 steps
Let g(x) = ∫₀^x (t² − 4) dt. Where is g increasing, where are its extremes, where does it bend, and what is g(2)?

Open this problem in the tool
Show solution
1. g′(x) = x² − 4 by part 1, positive outside [−2, 2] and negative inside it:
  
  $g^{'} (x) = x^{2} - 4 : g^{'} > 0 ⟺ ∣ x ∣ > 2, g^{'} < 0 ⟺ ∣ x ∣ < 2$
2. g″(x) = 2x changes sign at 0, so g bends there:
  
  $g^{''} (x) = 2 x = 0 ⟺ x = 0$
3. And g(2) by part 2:
  
  $g (2) = [\frac{t ^{3}}{3} - 4 t]_{0}^{2} = \frac{8}{3} - 8 = - \frac{16}{3}$
Answer

g rises on (−∞, −2] and [2, ∞) and falls on [−2, 2]; a maximum at −2, a minimum at 2, an inflection at 0; g(2) = −16/3 ≈ −5.3333. Every one of those came from the sign of t² − 4 and its slope, before g itself was written down.

$max : x = - 2, min : x = 2, g (2) = - \frac{16}{3} \approx - 5.3333$
The logarithm as an area 1 steps
Evaluate ∫₁^e dt/t.

Open this problem in the tool
Show solution
1. ln t is an antiderivative of 1/t on t > 0:
  
  $\int_{1}^{e} \frac{d t}{t} = [ln t]_{1}^{e} = ln e - ln 1 = 1 - 0$
Answer

1. Read backwards, this is the definition many books start from: ln x is the area under 1/t from 1 to x, and e is the number where that area reaches 1.

$\int_{1}^{e} \frac{d t}{t} = 1$

The slope of the area is the height of the curve, formula or not 🖖

Press e^(−t²): no formula. The page never uses an antiderivative to draw the lower curve: it adds up the area under e^−t² slice by slice. At x = 1 the area is 0.7468, and the lower curve is rising there at 0.3679 per unit, which is the height of the bell at 1. Move x anywhere and the two numbers stay equal. The reason fits in a sentence: widen the interval from x to x + h and the area grows by a strip h wide and about f(x) tall, so it grows at f(x) per unit of x. Nothing in that needs a formula. Liouville proved e^−t² has none, and the theorem hands it an antiderivative anyway: the area function itself.

Every area function is an antiderivative, and not the other way round 🖖

Press cos t from a = 2. Moving the start a slides the whole lower curve up or down without changing its shape, so its slope at x stays cos 1 = 0.5403 whatever a is. Every choice of a gives an antiderivative of cos t. The converse fails. sin t + 5 is an antiderivative of cos t too, and no area function equals it: an area function is 0 at its start, and sin t + 5 never drops below 4. ∫ₐˣ picks out the antiderivatives that pass through zero somewhere, and the theorem only ever needs one of them.

The area peaks where the curve crosses zero going down 🖖

Press sin t to π. The area is 2 and the lower curve is flat: before π the curve is above the axis and adds area, after π it is below and takes area away, so π is the top. Press t² − 4 to 2: the area is −5.3333 at a bottom, for the mirror reason. An exam asks for these from the graph of f alone, and the theorem is why that is enough: the area function rises where f is positive, falls where f is negative, peaks where f crosses from positive to negative, and bends where f has a peak or a trough of its own. The grey dots on the lower panel are its turning points.

Learning path

Applications of integration

References (1)

Both parts as the page states them: Theorem 5.4 (if f is continuous on [a, b] and F(x) = ∫ₐˣ f(t) dt, then F′(x) = f(x) on (a, b)) and Theorem 5.5 (∫ₐᵇ f(x) dx = F(b) − F(a) for any antiderivative F of a continuous f): G. Strang and E. Herman, Calculus Volume 1, OpenStax (2016), section 5.3, “The Fundamental Theorem of Calculus”.

By the 42Math team · Reviewed against published sources. Corrections welcome.