Definite Integral as Limit of a Sum

A definite integral of a function $f(x)$ on an interval $[a,b]$ is written $\int_a^b f(x)\,dx$, read as "the integral from $a$ to $b$ of $f(x)$ with respect to $x$." Here $a<b$ are real numbers and $f$ is defined (and, for this introduction, continuous and non-negative) on $[a,b]$.

Geometric meaning. $\int_a^b f(x)\,dx$ is defined as the area of the plane region bounded above by the curve $y=f(x)$, below by the X-axis, and on the sides by the vertical lines $x=a$ and $x=b$. If $g(x)$ is any primitive (antiderivative) of $f(x)$ — that is, $g'(x)=f(x)$ — then this area turns out to equal $g(b)-g(a)$. The rest of this section explains why that is true, by building the area up from thin strips and taking a limit.

Partitioning the interval. Divide $[a,b]$ into $n$ equal sub-intervals using the points

Each sub-interval $[x_r,x_{r+1}]$ determines a thin vertical strip of the region under the curve. The width of every strip is the same, $h=\dfrac{b-a}{n}$, so that $x_r=a+rh$ and, since $x_n=b$, we always have $nh=b-a$.

Approximating a strip by a rectangle. Figure 4.1 shows one such strip. Its true area (bounded above by the curved arc of $y=f(x)$) is approximated by the area of a rectangle with the same base $[x_r,x_{r+1}]$ and height $f(t_r)$, where $t_r$ is some point strictly between $x_r$ and $x_{r+1}$ chosen by the Mean Value Theorem (explained next). This rectangle's area is $(x_{r+1}-x_r)\cdot f(t_r)$.

Bringing in the Mean Value Theorem. If $g(x)$ is a primitive of $f(x)$, the Mean Value Theorem applied to $g$ on $[x_r,x_{r+1}]$ says there exists $t_r$ with $x_r<t_r<x_{r+1}$ such that

because $g'=f$. So the exact increase in $g$ across the strip equals exactly the approximate rectangle area built from $f(t_r)$ — the approximation is not a coincidence, it is forced by the Mean Value Theorem.

Summing all strips. Adding the rectangle areas for $r=0,1,\dots,n-1$ gives an approximating sum $S_n$ for the whole region:

The right-hand sum telescopes — every interior $g(x_r)$ cancels once as a subtrahend and once as a minuend — leaving only the first and last terms:

This is true for every $n$, so in particular

This limit — the limit, as the strips get infinitely thin, of the sum of thin-rectangle areas — is defined to be the definite integral:

The limit-of-sum working formula. In practice $t_r$ is taken to be the right end-point $a+rh$ of each strip (a legitimate choice since the Mean Value Theorem only guarantees some point in the strip, and for a continuous $f$ the limit is the same regardless of which point in each strip is used), so the formula actually used to evaluate an integral from first principles is

The word "integrate" literally means "to find the sum of" — reflecting exactly this construction. This technique underlies not just areas but also arc lengths, volumes of revolution, and similar "sum of infinitely many infinitesimal pieces" constructions met later.

Worked Example 1 — $\int_1^2(2x+5)\,dx$. Here $f(x)=2x+5$, $a=1,\ b=2$, so $h=\frac{1}{n}$ and $nh=1$. Then $f(a+rh)=f(1+rh)=2(1+rh)+5=7+2rh$. The sum becomes

As $n\to\infty$, $hn=1$ stays fixed and $1/n\to0$, so the limit is $7(1)+(1)^2(1+0)=8$. So $\int_1^2(2x+5)\,dx=8$.

Worked Example 2 — $\int_2^3 7^x\,dx$. Here $f(x)=7^x$, $a=2,b=3$, $h=1/n$. Then $f(a+rh)=7^{2+rh}=7^2\cdot 7^{rh}$, and the sum $\sum_{r=1}^n 7^{rh}$ is a finite geometric series with common ratio $7^h$ and $n$ terms, summing to $7^h\dfrac{(7^h)^n-1}{7^h-1}=\dfrac{7^h(7^{nh}-1)}{7^h-1}$. Multiplying by $h\cdot7^2$ and using $nh=1$, the sum is $49\cdot\dfrac{7^h(7-1)}{7^h-1}\cdot h= 49\cdot 6\cdot\dfrac{h}{7^h-1}\cdot 7^h$. As $h\to0$, $7^h\to1$ and $\dfrac{7^h-1}{h}\to\log 7$ (the defining limit of the derivative of $7^x$ at $0$), so $\dfrac{h}{7^h-1}\to\dfrac1{\log7}$. The limit is therefore $\dfrac{49\cdot6}{\log7}=\dfrac{294}{\log 7}$.

Worked Example 3 — $\int_0^4 (x-x^2)\,dx$. Here $f(x)=x-x^2$, $a=0,b=4$, $h=4/n$, and $f(rh)=rh-r^2h^2$. Splitting the sum,

Writing each in terms of the fixed quantity $nh=4$: $h^2n(n+1)=(nh)^2(1+1/n)\to 16$, and $h^3n(n+1)(2n+1)=(nh)^3(1+1/n)(2+1/n)\to 64\cdot2=128$. So the limit is $\tfrac{16}{2}-\tfrac{128}{6}=8-\tfrac{64}{3}=-\tfrac{40}{3}$.

Worked Example 4 — $\int_0^{\pi/2}\sin x\,dx$. Here $a=0,\ b=\pi/2,\ h=\dfrac{\pi/2}{n}$, so $nh=\pi/2$, and $f(rh)=\sin(rh)$. The sum $\sum_{r=1}^n \sin(rh)$ is a standard trigonometric sum; multiplying it by $2\sin\frac h2$ turns every term into a telescoping difference of cosines via the identity $2\sin A\sin B=\cos(A-B)-\cos(A+B)$, and collapses to

using $nh=\pi/2$ and $\cos(\pi/2+\theta)=-\sin\theta$. So $\displaystyle\sum_{r=1}^n\sin(rh)=\frac{\cos\frac h2+\sin\frac h2}{2\sin\frac h2}$, and the original sum is

As $n\to\infty$, $h\to0$, so $\cos\frac h2\to1$, $\sin\frac h2\to0$, and $\dfrac{h/2}{\sin(h/2)}\to1$ (the standard $\theta/\sin\theta\to1$ limit). The whole expression $\to (1+0)\cdot1=1$. Hence $\int_0^{\pi/2}\sin x\,dx=1$.