The exact integral

The more bars we take for the approximation on an integral, the thinner the bars will be, because they all have to fit into the integration limits $a$ and $b$ . This means we have to add more bar areas, but the approximation gets more accurate. You can check this out with the Geogebra app below for the integral

\int_0^6 ((x-3)^2-2.25)\, dx

Increase the number of bars using the slider.

Open in GeoGebra

Observe how the sum of bars approaches a specific value (here $4.5$ ) if we use more and more bars, that is, if we let $n$ approach infinity. This specific value is the exact value of the integral:

\int_0^6 ((x-3)^2-2.25)\, dx = 4.5

Thus, the exact integral is obtained by a limit process:

Theorem 1

For increasingly bigger bar numbers, the Rieman sum (sum of bars) converges towards the integral. This is written like this:

Equation 1

\int_a^b f(x)\, dx = \lim_{n\rightarrow \infty} \sum_{k=1}^n f(x_n)\Delta x

The Rieman sum converges towards the integral for bar number approaching infinity.

Insight

We can think of $\int_a^b f(x)\, dx$ as a Rieman sum of "infinitely many, infinitely thin bars".

\underbrace{\int_a^b}_{sum} \underbrace{\overbrace{f(x)}^{height}\, \overbrace{dx}^{width}}_{bar\, area}

However, unlike $\Delta x$ , there is no value associated with $dx$ ... . You can think of it as a number bigger than $0$ than zero, but smaller than any positive number. Clearly such a number does not exist, and thus $dx$ is more a symbol for the width of incredibly thin bars.

Here is an example.

Example 1

Determine the integral $\int_0^3 3x\, dx$ :

Approximate $\int_0^3 3x\, dx$ using $n$ bars and find a formula for the sum of the $n$ bars. Hint: you need one of the famous sums, see section 25.
Let $n$ approach infinity to find the exact value $\int_0^3 3x\,dx$ .
Determine the exact area $A_f$ by other means and compare with the result above.

Solution

For $n$ bars it is

\Delta x = \frac{3-0}{n}=\frac{3}{n}

We also have

x_1=\frac{3}{n}, x_2=\frac{6}{n}, x_3=\frac{9}{n}, ..., x_n=\frac{3n}{n}=3

f(x_1)=\frac{9}{n}, f(x_2)=\frac{18}{n}, f(x_3)=\frac{27}{n}..., f(x_n)=\frac{9n}{n}=9

Approximating $\int_0^3 3x\, dx$ with $n$ bars, we get $\begin{array}{lll} \int_0^3 3x\, dx &\approx & \sum_{k=1}^n f(x_k)\Delta x \\[0.1em] & = & \frac{9}{n}\cdot\frac{3}{n}+\frac{18}{n}\cdot\frac{3}{n}+\frac{27}{n}\cdot\frac{3}{n}+...+\frac{9n}{n}\cdot \frac{3}{n}\\[0.1em] &=& \frac{27}{n^2}+\frac{54}{n^2}+\frac{81}{n^2}+...+\frac{27n}{n^2}\\[0.1em] &=& \frac{1}{n^2}(27+54+81+...+27n)\\[0.1em] &=& \frac{27}{n^2}(\underbrace{1+2+3+...+n}_{\frac{n(n+1)}{2}})\\[0.1em] &=& \frac{27}{n^2}\frac{n^2+n}{2}\\[0.1em] &=& \frac{27}{2}\frac{n^2+n}{n^2}\\[0.1em] &=& 13.5\cdot (\frac{n^2}{n^2}+\frac{n}{n^2})\\[0.1em] &=& 13.5\cdot (1+\frac{1}{n})\\[0.1em] \end{array}$ Note that we have used the sum formula $1+2+...+n=\frac{n(n+1)}{2}$ (see section 2, Exercise 2).
Now we see that if $n\rightarrow \infty$ we get $\int_0^3 3x\, dx = 13.5\cdot (1+0)= \underline{13.5}$ This is also the area $A_f$ .
The region is a triangle with base $3$ and height $f(3)=9$ . Thus, the area is $\frac{3\cdot 9}{2} = \underline{13.5}$

Exercise 1

Determine

\int_0^1 x^2\, dx

Hint: You need the formula for summing the first $n$ square numbers, see last exercise in the sigma-notation chapter.

Solution

Using $n$ bars in the Rieman sum, we have for the bar width

\Delta x = \frac{1-0}{n}=\frac{1}{n}

and for the right edges of each bar the positions

x_1=\frac{1}{n}, x_2=\frac{2}{n}, x_3=\frac{3}{n}, ..., x_n=\frac{n}{n}=1

and the bar heights are

f(x_1)=\frac{1^2}{n^2}, f(x_2)=\frac{2^2}{n^2}, f(x_3)=\frac{3^2}{n^2}, ..., f(x_n)=\frac{n^2}{n^2}

Thus, we have

\begin{array}{lll} \int_0^1 x^2\,dx &\approx& \sum_{k=1}^n f(x_k)\Delta x \\[0.1em] & =& f(x_1)\Delta x+f(x_2)\Delta x +f(x_3)\Delta x + ...+ f(x_n)\Delta x \\[0.1em] & = & \left(\frac{1}{n}\right)^2 \frac{1}{n}+\left(\frac{2}{n}\right)^2 \frac{1}{n}+\left(\frac{3}{n}\right)^2 \frac{1}{n}+...+\left(\frac{n}{n}\right)^2 \frac{1}{n}\\[0.1em] & = & \frac{1^2}{n^3} + \frac{2^2}{n^3}+\frac{3^2}{n^3}+...+\frac{n^2}{n^3}\\[0.1em] & = & \frac{1}{n^3} (\underbrace{1^2 +2^2+3^2+...+n^2}_{\frac{n(n+1)(2n+1)}{6}})\\[0.1em] &=& \frac{n(n+1)(2n+1)}{6 n^3}\\[0.1em] &=& \frac{(n^2+n)(2n+1)}{6 n^3}\\[0.1em] &=& \frac{2n^3+3n^2+n}{6 n^3}\\[0.1em] &=& \frac{2n^3}{6 n^3}+\frac{3n^2}{6 n^3}+\frac{n}{6 n^3}\\[0.1em] &=& \frac{1}{3}+\frac{1}{2 n}+\frac{1}{6 n^2} \end{array}

Note that we have used the (formula ) for finding the first $n$ square numbers:

1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}

Finally, $n\rightarrow \infty$ we get

\int_0^1 x^2\,dx =\frac{1}{3}+0+0=\underline{\frac{1}{3}}

Finally, we state some properties of the integral, which sometimes are useful.

Theorem 2

Consider two functions $f$ and $g$ , and two constants $s$ and $t$ . Integrating over the weighted sum is the same as taking the weighted sum of the integrals:

Equation 2

\int_a^b (sf(x)\pm tg(x) )\, dx = s\int_a^b f(x)\, dx \pm t\int_a^b g(x)\, dx

Weighted sum of integrals

Proof

We show it just for the $+$ version, the $-$ version is similar. For every number of bars $n$ we can approximate the integral using a Reiman sum:

\begin{array}{lll} \int_a^b (sf(x) + tg(x) )\, dx &\approx& \sum_{k=1}^n (sf(x_k)+tg(x_k))\Delta x\\[1em] &=& (sf(x_1)+tg(x_1)+\dots+sf(x_1)+tg(x_1))\Delta x\\[1em] &=& s(f(x_1)\Delta x+\dots+f(x_n)\Delta x)+t(g(x_1)\Delta x+\dots+g(x_n)\Delta x)\\[1em] &=& s\sum_{k=1}^n f(x_k) \Delta x + t\sum_{k=1}^n g(t_k)\Delta x \end{array}

Letting $n$ move towards infinity, we have $\sum \rightarrow \int$ and $\approx \rightarrow =$ , so we get indeed

\int_a^b (sf(x)\pm tg(x) )\, dx = s\int_a^b f(x)\, dx \pm t\int_a^b g(x)\, dx

Example 2

$\int_0^1 2x^2\,dx = 2\int_0^1 x^2\, dx$
$\int_0^1 (x^2+x^3)\,dx = \int_0^1 x^2 + \int_0^1 x^3\, dx$
$\int_0^1 (x^2-3x^3)\,dx= \int_0^1 x^2 - 3\int_0^1 x^3\, dx$
$\int_0^1 (-4x^2+3x^3)\,dx= -4\int_0^1 x^2 + 4\int_0^1 x^3\, dx$