Area and antiderivatives (the FTC intuition)

Let $S(x)$ be the area between $y=f(x)$ and the $x$-axis from 0 to $x$. Increasing $x$ a little adds a thin strip of height $f(x)$, so the rate of change of $S(x)$ is $f(x)$.

For $y=6$, $S(x)=6x$; for $y=6x$, $S(x)=3x^2$; differentiate either and you get the function back. So finding an area turns into finding a function whose derivative is $f$, an antiderivative. That is the content of the Fundamental Theorem of Calculus.

Examples

1. 01What is the area S(x) under this line from 0 to x?$\displaystyle y=6$AnswerHide

   Answer$\displaystyle 6x$
2. 02S(x) is the area from 0 to x. What is S(4)?$\displaystyle y=6x, \quad S(x)=3x^2$AnswerHide

   Answer$\displaystyle 48$
3. 03S(x) is the area from 0 to x. What is S(3)?$\displaystyle y=10x, \quad S(x)=5x^2$AnswerHide

   Answer$\displaystyle 45$
4. 04$S(x)$ is the area from $0$ to $x$. What goes in □?$\displaystyle y=f(x), \quad S'(x)=\square$AnswerHide

   Answer$\displaystyle f(x)$
5. 05What is the area S(x) under this line from 0 to x?$\displaystyle y=5$AnswerHide

   Answer$\displaystyle 5x$
6. 06S(x) is the area from 0 to x. Differentiate:$\displaystyle S(x)=6x$AnswerHide

   Answer$\displaystyle 6$

FAQ

Q1Why is S(0)=0?Show answerHide

A

An interval of width $0$ has area $0$. In terms of an antiderivative $F(x)$, $S(x)=F(x)-F(0)$: subtracting the constant $F(0)$ makes $S(0)=0$.

Practice this pattern in the app, 5 problems a day.

Published 2026-09-04 · Updated 2026-09-05