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

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.

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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.
$\frac{A(x+h)-A(x)}{h}=\frac{1}{h}\int_{x}^{x+h}f(t)\,dt\ \longrightarrow\ f(x)$
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.
$A(x)=F(x)+C,\quad A(a)=0\ \Rightarrow\ C=-F(a)\ \Rightarrow\ \int_{a}^{b}f=F(b)-F(a)$

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.
  
  $\frac{d}{dx}\int_{1}^{x}\sqrt{1+t^{3}}\,dt=\sqrt{1+x^{3}}$
Answer

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

$\sqrt{1+x^{3}}$
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:
  
  $\frac{d}{dx}\int_{0}^{x^{2}}\cos t\,dt=\cos\left(x^{2}\right)\cdot2x$
2. Check by doing the integral first:
  
  $\int_{0}^{x^{2}}\cos t\,dt=\sin\left(x^{2}\right)\ \Rightarrow\ \frac{d}{dx}\sin\left(x^{2}\right)=2x\cos\left(x^{2}\right)$
Answer

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

$2x\cos\left(x^{2}\right)$
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:
  
  $\int_{x}^{2x}t^{2}\,dt=\int_{0}^{2x}t^{2}\,dt-\int_{0}^{x}t^{2}\,dt$
2. Differentiate each piece, with the chain rule on the upper one:
  
  $\frac{d}{dx}=(2x)^{2}\cdot2-x^{2}=8x^{2}-x^{2}=7x^{2}$
3. Check by doing the integral first:
  
  $\int_{x}^{2x}t^{2}\,dt=\frac{8x^{3}-x^{3}}{3}=\frac{7x^{3}}{3}\ \Rightarrow\ 7x^{2}$
Answer

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

$7x^{2}$
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:
  
  $\int_{0}^{\pi}\sin x\,dx=\Bigl[-\cos x\Bigr]_{0}^{\pi}=-\cos\pi+\cos0=1+1$
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.

$\int_{0}^{\pi}\sin x\,dx=2$
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:
  
  $g'(x)=x^{2}-4:\quad g'>0\ \iff\ |x|>2,\qquad g'<0\ \iff\ |x|<2$
2. g″(x) = 2x changes sign at 0, so g bends there:
  
  $g''(x)=2x=0\iff x=0$
3. And g(2) by part 2:
  
  $g(2)=\Bigl[\tfrac{t^{3}}{3}-4t\Bigr]_{0}^{2}=\tfrac{8}{3}-8=-\tfrac{16}{3}$
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.

$\max:\ x=-2,\quad\min:\ x=2,\quad g(2)=-\tfrac{16}{3}\approx-5{.}3333$
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:
  
  $\int_{1}^{e}\frac{dt}{t}=\Bigl[\ln t\Bigr]_{1}^{e}=\ln e-\ln1=1-0$
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.

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

Supplementary problems, answers only

Find d/dx ∫_x² e^t² dt.

Show answer

−e^x². The variable is the lower limit, so the sign changes.

$-e^{x^{2}}$
Find d/dx ∫₀^{sin x} √(1 − t²) dt for −π/2 < x < π/2.

Show answer

√(1 − sin²x) · cos x = cos²x, because cos x is positive on that interval.

$\cos^{2}x$
Evaluate ∫₋₁² (x² − 1) dx. Is the result the area between the curve and the x-axis?

Show answer

0, and the area is not 0 but 8/3: the 4/3 below the axis on [−1, 1] cancels the 4/3 above it on [1, 2].

$\int_{-1}^{2}\left(x^{2}-1\right)dx=0,\qquad A=\tfrac{8}{3}$
Find f if ∫₀^x f(t) dt = x² sin x for every x.

Show answer

f(x) = 2x sin x + x² cos x, by differentiating both sides.

$f(x)=2x\sin x+x^{2}\cos x$
Find the tangent line to G(x) = ∫₁^x √(t + 3) dt at x = 1.

Show answer

y = 2x − 2. G(1) = 0 because the interval has no width, and G′(1) = √4 = 2.

$y=2x-2$

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 7 Raven Room team · Reviewed against published sources. Corrections welcome.