Definite Integrals

To measure a curved region, divide it into narrow vertical strips. Each strip has approximate area height × width. As the widths decrease, the sum approaches the exact area. Below the x-axis, negative heights give negative contributions.

A definite integral records this signed total between two limits. In $\int_a^b f(x)\,dx$, $a$ is the starting input and $b$ the ending input. For area, normally $a<b$. The $dx$ corresponds to strip widths along the x-axis.

Why does reversing differentiation find area? Increasing the right boundary by a small amount adds a strip of height $f(x)$. As its width tends to zero, the rate at which the accumulated area changes is $f(x)$. So a function with derivative $f(x)$ gives the accumulated total, up to a constant.

An antiderivative $F(x)$ is a function whose derivative is $f(x)$. Subtracting its starting value leaves just the contribution between the limits:

The square brackets mean evaluate at the upper limit, then subtract the value at the lower limit. The constant cancels: $(F(b)+c)-(F(a)+c)=F(b)-F(a)$.

Use this calculation directly when the integrand is defined and continuous throughout the interval: no missing values or breaks. Undefined endpoints and infinite limits need a limiting calculation, taught later in this chapter.

Equal limits give zero. Exchanging the limits reverses the sign of $F(b)-F(a)$, not the geometric region.

Split an interval into adjacent parts and add their integrals to recover the original total. A constant multiplying the integrand multiplies that total by the same constant.

A calculator's numerical integral is a useful check. Write the antiderivative and the substituted limits when working is required; a decimal alone does not show those steps or establish an exact fraction.