Concept of Definite Integral

Related calculators: Definite and Improper Integral Calculator , Riemann Sum Calculator for a Function

In Area Problem note we saw that limit of the form $$$\lim_{{{n}\to\infty}}{f{{\left({{x}_{{i}}^{{\star}}}\right)}}}\Delta{x}$$$ arises when we compute an area.

It turns out that this same type of limit occurs in a wide variety of situations even when $$${f{}}$$$ is not necessarily a positive function.

Thus, this limit has special name and notation.

Definition of a Definite Integral. If $$$f$$$ is a continuous function defined for $$${a}\le{x}\le{b}$$$, we divide the interval $$${\left[{a},{b}\right]}$$$ into $$${n}$$$ subintervals of equal width $$$\Delta{x}=\frac{{{b}-{a}}}{{n}}$$$. We let $$${x}_{{0}}{\left(={a}\right)},{x}_{{1}},{x}_{{2}},\ldots,{x}_{{n}}{\left(={b}\right)}$$$ be the endpoints of these subintervals and we choose sample points $$${{x}_{{1}}^{{\star}}},{{x}_{{2}}^{{\star}}},\ldots,{{x}_{{n}}^{{\star}}}$$$ in these subintervals, so $$${{x}_{{i}}^{{\star}}}$$$ lies in the i-th subinterval $$${\left[{x}_{{{i}-{1}}},{x}_{{i}}\right]}$$$. Then the definite integral of $$${f{}}$$$ from $$${a}$$$ to $$${b}$$$ is $$${\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}=\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{n}}}{f{{\left({{x}_{{i}}^{{\star}}}\right)}}}\Delta{x}$$$.

As with indefinite integral case $$${f{{\left({x}\right)}}}$$$ is called integrand.

$$${a}$$$ is lower limit and $$${b}$$$ is upper limit.

Note, that definite integral is a number, it doesn't depend on $$${x}$$$. We can choose another variable: $$${\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}={\int_{{a}}^{{b}}}{f{{\left({t}\right)}}}{d}{t}={\int_{{a}}^{{b}}}{f{{\left({u}\right)}}}{d}{u}$$$.

The sum $$${\sum_{{{i}={1}}}^{{n}}}{f{{\left({{x}_{{i}}^{{\star}}}\right)}}}\Delta{x}$$$ is called Riemann sum.

If sample points $$${{x}_{{i}}^{\star}}$$$ are left endpoints then sum is called Left Riemann Sum, if right endpoints then sum is called Right Riemann Sum.

Now let's see what will be if $$${f{}}$$$ can take both positive and negative values.net area

If $$${f{}}$$$ takes on both positive and negative values, then the Riemann sum is the sum of the areas of the rectangles that lie above the x-axis minus sum of areas of the rectangles that lie below the x-axis.

Thus, definite integral is net area: area above x-axis minus area below x-axis:

$$${\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}={S}_{{1}}-{S}_{{2}}+{S}_{{3}}+{S}_{{4}}-{S}_{{5}}+{S}_{{6}}$$$.

Note, that although we defined $$${\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}$$$ by dividing interval $$${\left[{a},{b}\right]}$$$ into $$${n}$$$ intervals of equal width, it is sometimes advantageous to work with intervals of unequal width. If the subinterval widths are $$$\Delta{x}_{{1}},\Delta{x}_{{2}},\ldots,\Delta{x}_{{n}}$$$ then we have to ensure that all these widths approach 0 in the limiting process. This happens if the largest width, $$$\max\Delta{x}_{{i}}$$$, approaches 0. So in this case definition of integral becomes $$${\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}=\lim_{{\max\Delta{x}_{{i}}\to{0}}}{\sum_{{{i}={1}}}^{{n}}}{f{{\left({{x}_{{i}}^{{\star}}}\right)}}}\Delta{x}_{{i}}$$$.

Now let's go through a couple of examples.

Example 1. Express $$$\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{n}}}{\left[{3}{{x}_{{i}}^{{2}}}-{\cos{{\left({{x}_{{i}}^{{5}}}\right)}}}\right]}\Delta{x}$$$ as an integral on the interval $$${\left[{0},\pi\right]}$$$.

Comparing the given limit with the limit in definition of integral, we see that they will be identical if we choose $$${f{{\left({x}\right)}}}={3}{{x}}^{{2}}-{\cos{{\left({{x}}^{{5}}\right)}}}$$$ and $$${{x}_{{i}}^{{\star}}}={x}_{{i}}$$$ (so the sample points are right endpoints). We are given that $$${a}={0}$$$ and $$${b}=\pi$$$, therefore $$$\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{n}}}{\left[{3}{{x}_{{i}}^{{2}}}-{\cos{{\left({{x}_{{i}}^{{5}}}\right)}}}\right]}\Delta{x}={\int_{{0}}^{\pi}}{\left({3}{{x}}^{{2}}-{\cos{{\left({{x}}^{{5}}\right)}}}\right)}{d}{x}$$$.

It is very important to recognize limits of sums as integrals. In general, if $$$\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{n}}}{f{{\left({{x}_{{i}}^{{\star}}}\right)}}}\Delta{x}={\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}$$$ then we replace $$$\lim\sum$$$ by $$$\int$$$, $$${{x}_{{i}}^{{\star}}}$$$ by $$${x}$$$ and $$$\Delta{x}$$$ by $$${d}{x}$$$.

Example 2. Evaluate $$${\int_{{0}}^{{2}}}{\left({3}{{x}}^{{2}}-{{x}}^{{3}}\right)}{d}{x}$$$.

Divide interval $$${\left[{0},{2}\right]}$$$ into $$${n}$$$ subintervals of width $$$\Delta{x}=\frac{{{2}-{0}}}{{n}}=\frac{{2}}{{n}}$$$.

Since we can use any point within subinterval, let's use right endpoints.

Right endpoint of i-th subinterval is $$$\frac{{{2}{i}}}{{n}}$$$, so

$$${\int_{{0}}^{{2}}}{\left({3}{{x}}^{{2}}-{{x}}^{{3}}\right)}{d}{x}=\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{n}}}{f{{\left(\frac{{{i}}}{{n}}\right)}}}\frac{{2}}{{n}}=\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{n}}}{\left({3}{{\left(\frac{{{2}{i}}}{{n}}\right)}}^{{2}}-{{\left(\frac{{{2}{i}}}{{n}}\right)}}^{{3}}\right)}\frac{{2}}{{n}}=\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{n}}}{\left({24}\frac{{{i}}^{{2}}}{{{n}}^{{2}}}-{16}\frac{{{i}}^{{3}}}{{{n}}^{{3}}}\right)}\frac{{1}}{{n}}=$$$

$$$=\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{n}}}{\left(\frac{{24}}{{{n}}^{{3}}}{{i}}^{{2}}-\frac{{16}}{{{n}}^{{4}}}{{i}}^{{3}}\right)}=\lim_{{{n}\to\infty}}{\left({\sum_{{{i}={1}}}^{{n}}}{\left(\frac{{24}}{{{n}}^{{3}}}{{i}}^{{2}}\right)}-{\sum_{{{i}={1}}}^{{n}}}{\left(\frac{{16}}{{{n}}^{{4}}}{{i}}^{{3}}\right)}\right)}=\lim_{{{n}\to\infty}}{\left(\frac{{24}}{{{n}}^{{3}}}{\sum_{{{i}={1}}}^{{n}}}{{i}}^{{2}}-\frac{{16}}{{{n}}^{{4}}}{\sum_{{{i}={1}}}^{{n}}}{{i}}^{{3}}\right)}=$$$

On this stage we need the following two formulas:

$$${\sum_{{{i}={1}}}^{{n}}}{{i}}^{{2}}=\frac{{{n}{\left({n}+{1}\right)}{\left({2}{n}+{1}\right)}}}{{6}}$$$.

$$${\sum_{{{i}={1}}}^{{n}}}{{i}}^{{3}}={{\left(\frac{{{n}{\left({n}+{1}\right)}}}{{2}}\right)}}^{{2}}$$$.

So, $$${\int_{{0}}^{{2}}}{\left({3}{{x}}^{{2}}-{{x}}^{{3}}\right)}{d}{x}=\lim_{{{n}\to\infty}}{\left(\frac{{24}}{{{n}}^{{3}}}\frac{{{n}{\left({n}+{1}\right)}{\left({2}{n}+{1}\right)}}}{{6}}-\frac{{16}}{{{n}}^{{4}}}{{\left(\frac{{{n}{\left({n}+{1}\right)}}}{{2}}\right)}}^{{2}}\right)}=$$$

$$$=\lim_{{{n}\to\infty}}{\left(\frac{{{4}{\left({n}+{1}\right)}{\left({2}{n}+{1}\right)}}}{{{n}}^{{2}}}-\frac{{{4}{{\left({n}+{1}\right)}}^{{2}}}}{{{n}}^{{2}}}\right)}=$$$

$$$=\lim_{{{n}\to\infty}}{\left({4}{\left({1}+\frac{{1}}{{n}}\right)}{\left({2}+\frac{{1}}{{n}}\right)}-{4}{\left({1}+\frac{{2}}{{n}}+\frac{{1}}{{{n}}^{{2}}}\right)}\right)}={4}{\left({1}+{0}\right)}{\left({2}+{0}\right)}-{4}{\left({1}+{2}\cdot{0}+{0}\right)}={4}$$$

So, $$${\int_{{0}}^{{2}}}{\left({3}{{x}}^{{2}}-{{x}}^{{3}}\right)}{d}{x}={4}$$$.

Example 3. Evaluate the following integral by interpreting it in terms of areas: $$${\int_{{-{2}}}^{{2}}}\sqrt{{{4}-{{x}}^{{2}}}}{d}{x}$$$.

Since $$${f{{\left({x}\right)}}}=\sqrt{{{4}-{{x}}^{{2}}}}\ge{0}$$$, we can interpret this integral as the area under the curve $$${y}=\sqrt{{{4}-{{x}}^{{2}}}}$$$ from -2 to 2.

Squaring both sides give $$${{y}}^{{2}}={4}-{{x}}^{{2}}$$$ then $$${{x}}^{{2}}+{{y}}^{{2}}={4}$$$ and the required area is area of semicircle with radius 2.

Therefore, $$${\int_{{-{2}}}^{{2}}}\sqrt{{{4}-{{x}}^{{2}}}}{d}{x}=\frac{{1}}{{2}}\pi\cdot{{\left({2}\right)}}^{{2}}={2}\pi$$$.

Example 4. Evaluate the following integral by interpreting it in terms of areas: $$${\int_{{-{{1}}}}^{{2}}}{\left({2}{\left|{x}\right|}-{1}\right)}{d}{x}$$$.example of net area

First we draw graph of the function $$${y}={2}{\left|{x}\right|}-{1}$$$ on interval $$${\left[-{1},{2}\right]}$$$.

Recall that definite integral is net area: $$${\int_{{-{1}}}^{{2}}}{\left({2}{\left|{x}\right|}-{1}\right)}{d}{x}={S}_{{1}}-{S}_{{2}}-{S}_{{3}}+{S}_{{4}}$$$.

Now area $$${S}_{{1}}$$$ is area of right-angled triangle with legs $$$\frac{{1}}{{2}}$$$ and 1, so $$${S}_{{1}}=\frac{{1}}{{2}}\cdot\frac{{1}}{{2}}\cdot{1}=\frac{{1}}{{4}}$$$.

Similarly, $$${S}_{{2}}=\frac{{1}}{{2}}\cdot\frac{{1}}{{2}}\cdot{1}=\frac{{1}}{{4}}$$$, $$${S}_{{3}}=\frac{{1}}{{2}}\cdot\frac{{1}}{{2}}\cdot{1}=\frac{{1}}{{4}}$$$ and $$${S}_{{4}}=\frac{{1}}{{2}}\cdot\frac{{3}}{{2}}\cdot{3}=\frac{{9}}{{4}}$$$.

So, $$${\int_{{-{1}}}^{{2}}}{\left({2}{\left|{x}\right|}-{1}\right)}{d}{x}=\frac{{1}}{{4}}-\frac{{1}}{{4}}-\frac{{1}}{{4}}+\frac{{9}}{{4}}={2}$$$.