The Area Function's Derivative Is the Function

Measure the slope of the accumulated-area function. It matches the original function exactly.

Area grows at a rate equal to the height at that point. Nudge the right edge and a thin strip attaches. The strip's area is height times the nudge.

For example, filling a tank with 💧 water, the level rises at the rate the water pours in. Differentiate what has accumulated and you recover what is pouring.

A(x)=∫axf(t) dt  ⇒  A′(x)=f(x)A(x) = \int_a^x f(t)\,dt \;\Rightarrow\; A'(x) = f(x)

Part of these learning paths

Examples & Exercises

Show A(x)=∫0xt2 dt=x33A(x) = \int_0^x t^2\,dt = \frac{x^3}{3} and verify A′(x)=x2A'(x) = x^2.

A(x)=x33,A′(x)=x2A(x) = \frac{x^3}{3}, \quad A'(x) = x^2
x=1.5x = 1.5: the numerical derivative matches f(1.5)=2.25f(1.5) = 2.25.
Show proof

  1. By the power rule, ∫0xt2 dt=[t33]0x=x33\int_0^x t^2\,dt = \left[\frac{t^3}{3}\right]_0^x = \frac{x^3}{3}.

  2. Differentiating gives A′(x)=x2=f(x)A'(x) = x^2 = f(x), matching the first form of the Fundamental Theorem.

Compute

Compute ∫012x dx\int_0^1 2x\,dx via an antiderivative.

∫012x dx=[x2]01=1\int_0^1 2x\,dx = \left[x^2\right]_0^1 = 1
The area up to x=1x = 1 is 1.
Show proof

  1. An antiderivative of 2x2x is x2x^2 (since (x2)′=2x(x^2)' = 2x).

  2. By the second form of the theorem, ∫012x dx=12−02=1\int_0^1 2x\,dx = 1^2 - 0^2 = 1.

Answer

11

Find ddx∫xx2sin⁡t dt\frac{d}{dx}\int_x^{x^2} \sin t\,dt with the Fundamental Theorem and the chain rule.

ddx∫xx2sin⁡t dt=2xsin⁡(x2)−sin⁡x\frac{d}{dx}\int_x^{x^2}\sin t\,dt = 2x\sin(x^2) - \sin x
On the area function of sin⁡\sin, at x≈1.01x \approx 1.01.
Show proof

  1. Put F(u)=∫0usin⁡t dtF(u) = \int_0^u \sin t\,dt; the integrand sin⁡\sin is continuous, so the continuity hypothesis of the first form of the Fundamental Theorem holds, and it gives F′(u)=sin⁡uF'(u) = \sin u.

  2. The integral is F(x2)−F(x)F(x^2) - F(x), so the chain rule gives 2x F′(x2)−F′(x)=2xsin⁡(x2)−sin⁡x2x\,F'(x^2) - F'(x) = 2x\sin(x^2) - \sin x.

11. Integration