The Fundamental Theorem of Calculus

The area function

Take the curve $y=x^2$y = x² and shade the region under it from x = 0 up to some point x. Call the shaded area A(x). It is a function of x: move the right edge and the area changes. From area under a curve, $A\left(2\right)=\frac{2^3}{3}=\frac{8}{3}$A(2) = 2³/3 = 8/3, and $A\left(3\right)=\frac{3^3}{3}=9$A(3) = 3³/3 = 9.

Push the right edge a little further, from x to x + h. The shaded region gains a thin strip, h wide, standing between x and x + h.

The area under $y=x^2$y = x² from 0 up to x = 2, which is $A\left(2\right)=\frac{8}{3}$A(2) = 8/3. Moving the right edge to the right adds a strip to the shaded region.

How fast the area grows

The new strip is h wide, and its height is close to f(x), the height of the curve at the edge. So the area grows by about f(x) × h: $A\left(x+h\right)-A\left(x\right)\approx f\left(x\right)h$A(x + h) − A(x) ≈ f(x)h.

Divide by h. The rate at which the area grows, $\left(A\left(x+h\right)-A\left(x\right)\right)/h$(A(x + h) − A(x)) / h, is about f(x), and the thinner the strip, the closer its top comes to a flat line at height f(x). As h shrinks to 0 the approximation becomes exact: A'(x) = f(x). The area function differentiates to the curve’s own height.

Check it at x = 2, where f(2) = 4. A strip 0.01 wide adds $A\left(2.01\right)-A\left(2\right)=\frac{2.01^3}{3}-\frac{8}{3}=0.0402003$A(2.01) − A(2) = 2.01³/3 − 8/3 = 0.0402003 to the area, and 0.0402003 ÷ 0.01 = 4.02, close to 4.

A sliver under $y=x^2$y = x², starting at x = 1.2, where the height is 1.44. At width $\Delta x=0.2$Δx = 0.2 the rectangle has area 1.44 × 0.2 = 0.288, against the 0.339 of the strip under the curve. Narrow it to 0.01: the strip becomes a rectangle 1.44 tall, so the area grows at the rate 1.44 at x = 1.2.

Areas from antiderivatives

So A is an antiderivative of f. Now take any antiderivative F of f. A and F have the same derivative, so their graphs have the same gradient at every x, and one is the other slid up or down: A(x) = F(x) + c for some constant c.

The area from a to a is 0, so 0 = F(a) + c, which gives c = −F(a). Then the area from a to b is A(b) = F(b) − F(a). This is the fundamental theorem of calculus: the integral from a to b of f(x) dx is F(b) − F(a), for any antiderivative F. It is written ${\int }_a^{b}f\left(x\right)dx={\left[F\left(x\right)\right]}_a^b=F\left(b\right)-F\left(a\right)$∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) − F(a).

Any antiderivative gives the same answer, because the constant cancels. With $F\left(x\right)=\frac{x^3}{3}+5$F(x) = x³/3 + 5, the area under $y=x^2$y = x² from 1 to 3 is $\left(9+5\right)-\left(\frac{1}{3}+5\right)=\frac{26}{3}$(9 + 5) − (1/3 + 5) = 26/3, the same as with $\frac{x^3}{3}$x³/3.

Check $\frac{26}{3}\approx 8.6666667$26/3 ≈ 8.6666667 by strips: 1000 midpoint rectangles from 1 to 3 add to 8.666666, within a millionth of it.

An area that is a triangle

The region under y = 2x from 0 to 3 is a triangle with base 3 and height 6, so its area is ½ × 3 × 6 = 9. An antiderivative of 2x is $x^2$x², and $3^2-0^2=9$3² − 0² = 9. In general the integral of 2x from 0 to b is $b^2$b².

The region under y = 2x from 0 to 3: a triangle with base 3 and height 6, area 9, which is $3^2-0^2$3² − 0².

The usual mistakes

Giving the height in place of the area. For y = 2x from 0 to 3, 6 is the height at x = 3; the area is 9.

Giving the width. 3 is the length of the interval from 0 to 3, not the area over it.

Using F(b) alone. That works only when F(a) = 0: the area under $y=x^2$y = x² from 1 to 3 is $9-\frac{1}{3}=\frac{26}{3}$9 − 1/3 = 26/3, not 9.

Subtracting the wrong way round. F(a) − F(b) gives $-\frac{26}{3}$−26/3 for the area under $y=x^2$y = x² from 1 to 3.

A boiler’s gas

In the application below, a boiler burns gas at a rate that changes through the morning. The gas burnt is the area under the rate curve, read off one antiderivative as a difference of two values, then split at a change of tariff.