# Definite Integral is Area

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## Theorem

Let $f : \closedint a b \to \R$ be Darboux integrable over $\closedint a b$.

Let $A$ be the total signed area between $\map f x$ and the $x$-axis, and between the lines $x = a$ and $x = b$.

Then $A$ equals the Darboux integral of $f$ over $\closedint a b$.

## Proof

### Lemma

Let $f : \closedint a b \to \R_{\mathop \ge 0}$ be non-negative and Darboux integrable over $\closedint a b$.

Let $G$ be the point set of $\tuple {x, y}$ such that $a \le x \le b$ and $0 \le y \le \map f x$.

Let $A$ be the area of $G$, which is colloquially called the **area under the curve**.

Then $A$ equals the Darboux integral of $f$ over $\closedint a b$.

$\Box$

Define $f^+$ and $f^-$ to be the positive and negative parts of $f$, respectively.

Let $A^+$ be the area under $\map {f^+} x$ on $\closedint a b$.

Let $A^-$ be the area under $\map {f^-} x$ on the same interval.

By Positive Part of Darboux Integrable Function is Integrable and its corollary for the negative part,
$f^+$ and $f^-$ are Darboux integrable over $\closedint a b$.

By the lemma, $A^+$ and $A^-$ are equal to the Darboux integral over their respective parts.

Therefore:

\(\ds A\) | \(=\) | \(\ds A^+ - A^-\) | Definition of Signed Area | |||||||||||

\(\ds \) | \(=\) | \(\ds \int_a^b \map {f^+} x \rd x - \int_a^b \map {f^-} x \rd x\) | Lemma | |||||||||||

\(\ds \) | \(=\) | \(\ds \int_a^b \paren {\map {f^+} x - \map {f^-} x} \rd x\) | Linear Combination of Integrals | |||||||||||

\(\ds \) | \(=\) | \(\ds \int_a^b \map f x \rd x\) | Difference of Positive and Negative Parts |

$\blacksquare$