# Concept of Definite Integral

## 関連する計算機: 明確で不適切な積分計算機 , 関数のリーマン和計算機

In the Area Problem note, we saw that the 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 the same type of limit occurs in a wide variety of situations even when ${f{}}$ is not necessarily a positive function.

Thus, this limit has a special name and notation.

Definition of 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 the sample points ${{x}_{{1}}^{{\star}}},{{x}_{{2}}^{{\star}}},\ldots,{{x}_{{n}}^{{\star}}}$ in these subintervals, so that ${{x}_{{i}}^{{\star}}}$ 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 the indefinite integral, ${f{{\left({x}\right)}}}$ is called the integrand.

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

Note that the 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 the Riemann sum.

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

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

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 the sum of the areas of the rectangles that lie below the x-axis.

Thus, the definite integral is the net area: the area above the x-axis minus the area below the 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 the 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, the 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 the 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 that the sample points are the 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 the limits of sums as integrals. In general, if $\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{n}}}{f{{\left({{x}_{{i}}^{{\star}}}\right)}}}\Delta{x}={\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}$, we replace $\lim\sum$ by $\int$, ${{x}_{{i}}^{{\star}}}$ by ${x}$ and $\Delta{x}$ by ${d}{x}$.

Once it is clear, let's move on.

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 the width $\Delta{x}=\frac{{{2}-{0}}}{{n}}=\frac{{2}}{{n}}$.

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

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

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

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

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

This is going to become clearer with some more practice.

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 gives ${{y}}^{{2}}={4}-{{x}}^{{2}}$; then, ${{x}}^{{2}}+{{y}}^{{2}}={4}$, and the required area is the area of the semicircle with the radius $2$.

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

Now, let's see our final example.

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

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

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

Now, the area ${S}_{{1}}$ is the area of a right-angled triangle with the 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, ${\int_{{-{1}}}^{{2}}}{\left({2}{\left|{x}\right|}-{1}\right)}{d}{x}=\frac{{1}}{{4}}-\frac{{1}}{{4}}-\frac{{1}}{{4}}+\frac{{9}}{{4}}={2}$.