The Area Function's Derivative Is the Function

Measure the slope of the area function $A(x) = \int_a^x f(t)\,dt$ — it equals $f(x)$ . Drag $x$ .

A(x) = \int_a^x f(t)\,dt \;\Rightarrow\; A'(x) = f(x)

Part of these learning paths

Show $A(x) = \int_0^x t^2\,dt = \frac{x^3}{3}$ and verify $A'(x) = x^2$ .

A(x) = \frac{x^3}{3}, \quad A'(x) = x^2

x = 1.5

: the numerical derivative matches

f(1.5) = 2.25

Show proof

By the power rule, $\int_0^x t^2\,dt = \left[\frac{t^3}{3}\right]_0^x = \frac{x^3}{3}$ .
Differentiating gives $A'(x) = x^2 = f(x)$ , matching the first form of the Fundamental Theorem.

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

\int_0^1 2x\,dx = \left[x^2\right]_0^1 = 1

The area up to

x = 1

is 1.

Show proof

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

\frac{d}{dx}\int_x^{x^2}\sin t\,dt = 2x\sin(x^2) - \sin x

On the area function of

\sin

, at

x \approx 1.01

Show proof

Put $F(u) = \int_0^u \sin t\,dt$ ; then $F'(u) = \sin u$ by the first form.
The integral is $F(x^2) - F(x)$ , so the chain rule gives $2x\,F'(x^2) - F'(x) = 2x\sin(x^2) - \sin x$ .