The Definite Integral

A definite integral has upper and lower limits and produces a number (not a function):

Why no $+C$? Suppose we used $F(x) + C$:

$[F(x) + C]_a^b = (F(b) + C) - (F(a) + C) = F(b) - F(a)$

The constants cancel! This is why definite integrals never include $+C$.

Example:

$\int_1^3 2x\,dx = [x^2]_1^3 = 3^2 - 1^2 = 9 - 1 = 8$

$\int_a^b f(x)\,dx = F(b) - F(a)$, evaluate antiderivative at top limit, subtract bottom limit; No $+C$ for definite integrals, the constants cancel when you subtract $F(a)$ from $F(b)$

Pause, copy the definite integral rule $\int_a^b f(x)\,dx = F(b) - F(a)$ and the reason there is no $+C$, the constants cancel in $F(b) - F(a)$, into your book.

We just saw that $\int_a^b f(x)\,dx = F(b) - F(a)$ computes a number. That raises a question: what does that number represent geometrically, and why does $\int_0^{2\pi}\sin x\,dx = 0$ even though $\sin x$ has non-zero area? This card answers it → the definite integral measures signed area: regions above the $x$-axis contribute positively, regions below contribute negatively.

The definite integral $\int_a^b f(x)\,dx$ equals the signed area between the curve and the $x$-axis:

• Where $f(x) > 0$ (above axis): area counts as positive
• Where $f(x) < 0$ (below axis): area counts as negative

Example: $\int_0^{2\pi} \sin x\,dx = [-\cos x]_0^{2\pi} = (-\cos 2\pi) - (-\cos 0) = (-1)-(-1) = 0$. The positive area from $0$ to $\pi$ exactly cancels the negative area from $\pi$ to $2\pi$.

If you want the total physical area (always positive), compute each region separately and add the absolute values:

$\int_0^{\pi}\sin x\,dx + \left|\int_{\pi}^{2\pi}\sin x\,dx\right| = 2 + 2 = 4$

Signed area: above $x$-axis = positive contribution; below = negative contribution; Definite integral gives net signed area; may be zero if positive and negative regions cancel

Pause, copy the signed-area rule (above $x$-axis = positive, below = negative; definite integral gives net signed area; split at $x$-intercepts and add absolute values for total physical area) into your book.

We just saw that the definite integral measures signed area, and that regions below the axis contribute negatively. That raises a question: are there algebraic shortcuts, ways to split, reverse, or factor definite integrals without computing from scratch? This card answers it → five key properties: same limits give zero, reversing limits negates, constant multiples factor out, the sum rule splits, and you can split the interval at any interior point.

$\int_a^a f = 0$ · $\int_a^b f = -\int_b^a f$ · $\int_a^b kf = k\int_a^b f$; $\int_a^b(f+g) = \int_a^b f + \int_a^b g$ (linearity of integration)

Pause, copy the five properties ($\int_a^a f = 0$; reversed limits negate; constant multiple; sum rule; interval splitting $\int_a^b + \int_b^c = \int_a^c$) into your book.