Calculus/Integration

Definition of the Integral

Figure 1

Figure 2

The rough idea of defining the area under the graph of f is to approximate this area with a finite number of rectangles. Since we can easily work out the area of the rectangles we get an estimate of the area under the graph. If we use a larger number of rectangles we expect a better approximation, and the limit as we approach an infinite number of rectangles will give the exact area.

Suppose first that f is positive and a<b. We pick an integer n and divide the interval [a,b] into n subintervals of equal width (see Figure 2). As the interval [a,b] has width b-a each subinterval has width $\mathrm{\Delta }x=\frac{b-a}{n}.$ We denote the endpoints of the subintervals by $x_0,x_1,\dots ,x_n$ so

$x_i=a+i\mathrm{\Delta }x\text{ for }i=0,1,\dots ,n.$

Figure 3

Now for each $i=1,\dots ,n$ pick a sample point $x_i^{*}$ in the interval $[x_{i-1},x_i]$ and consider the rectangle of height $f(x_i^{*})$ and width $\mathrm{\Delta }x$ (see Figure 3). The area of this rectangle is $f(x_i^{*})\mathrm{\Delta }x$. By adding up the area of all the rectangles for $i=1,\dots ,n$ we get that the area S is approximated by

$A_n=f(x_1^{*})\mathrm{\Delta }x+f(x_2^{*})\mathrm{\Delta }x+\cdots +f(x_n^{*})\mathrm{\Delta }x.$

A more convenient way to write this is with the summation notation as

$A_n=\sum _{i=1}^{n}f(x_i^{*})\mathrm{\Delta }x.$

For each number n we get a different approximation. As n gets larger the width of the rectangles gets smaller which yields a better approximation (see Figures 4 and 5). In the limit as $A_n$ as n tends to infinity we get the area of S.

Definition of the Definite Integral Suppose f is a continuous function on [a,b] and $\mathrm{\Delta }x=\frac{b-a}{n}$. Then the definite integral of f between a and b is $\underset{a}{\overset{b}{\int }}f(x)dx=\underset{n\to \mathrm{\infty }}{\lim }{A}_n=\underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^{n}f(x_i^{*})\mathrm{\Delta }x.$
where $x_i^{*}$ are any sample points in the interval $[x_{i-1},x_i]$.

It is a fact that if f is continuous on [a,b] then this limit always exists and does not depend on the choice of the points $x_i^{*}\in [x_{i-1},x_i]$. For instance they may be evenly spaced, or distributed ambiguously throughout the interval. The proof of this is technical and is beyond the scope of this section.

Notation When considering the expression $\underset{a}{\overset{b}{\int }}f(x)dx$ the function f is called the integrand and the interval [a,b] is the interval of integration. Also a is called the lower limit and b the upper limit of integration.

One important feature of this definition is that we also allow functions which take negative values. If f(x)<0 for all x then $f(x_i^{*})<0$ so $f(x_i^{*})\mathrm{\Delta }x<0$. So the definite integral of f will be strictly negative. More generally if f takes on both positive an negative values then $\underset{a}{\overset{b}{\int }}f(x)dx$ will be the area under the positive part of the graph of f minus the area under the graph of the negative part of the graph (see Figure 6). For this reason we say that $\underset{a}{\overset{b}{\int }}f(x)dx$ is the signed area under the graph.

Figure 6

Figure

The area spanned by $f(x)$

A geometrical proof that anti-derivative gives the area

Suppose we have an function F(x) which returns the area between x and some unknown point u. (Actually, u is the first number before x which satisfies F(u) = 0, but our solution is independent from u, so we won't bother ourselves with it.) We don't even know if something like F exists or not, but we're going to investigate what clue do we have if it does exists.

We can use F to calculate the area between a and b, for instance, which is obviously F(b)-F(a); F is something general. Now, consider a rather peculiar situation, the area bounded at $x$ and $x+\mathrm{\Delta }x$, in the limit of $\mathrm{\Delta }x\to 0$. Of course it can be calculated by using F, but we're looking for another solution this time. As the right border approaches the left one, the shape seems to be an infinitesimal rectangle, with the height of f(x) and width of $\mathrm{\Delta }x$. So, the area reads:

$\text{Infinitesimal area}=\underset{\mathrm{\Delta }x\to 0}{\lim }f(x)\mathrm{\Delta }x$

Of course, we could use F to calculate this area as well:

$\text{Infinitesimal area}=\underset{\mathrm{\Delta }x\to 0}{\lim }F(x+\mathrm{\Delta }x)-F(x)$

By combining these eqautions, we have

$\underset{\mathrm{\Delta }x\to 0}{\lim }F(x+\mathrm{\Delta }x)-F(x)=\underset{\mathrm{\Delta }x\to 0}{\lim }f(x)\mathrm{\Delta }x$

If we divide both sides by $\mathrm{\Delta }x$, we get

$\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{F(x+\mathrm{\Delta }x)-F(x)}{\mathrm{\Delta }x}=f(x)$

which is an interesting result, because the left-hand side is the derivative of F with respect to x. This remarkable result doesn't tell us what F itself is, however it tells us what the derivative of F is, and it is f.

Independence of Variable

It is important to notice that the variable x did not play an important role in the definition of the integral. In fact we can replace it with any other letter, so the following are all equal: $\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}f(t)dt=\underset{a}{\overset{b}{\int }}f(u)du=\underset{a}{\overset{b}{\int }}f(w)dw.$ Each of these is the signed area under the graph of f between a and b.

Left and Right Handed Riemann Sums

We could have decided to choose all our sample points $x_i^{*}$ to be on the right hand side of the interval $[x_{i-1},x_i]$ (see Figure 7). Then $x_i^{*}=x_i$ for all i and the approximation that we called $A_n$ for the area becomes

$A_n=\sum _{i=1}^{n}f(x_i)\mathrm{\Delta }x.$

This is called the right-handed Riemann sum, and the integral is the limit

$\underset{a}{\overset{b}{\int }}f(x)dx=\underset{n\to \mathrm{\infty }}{\lim }{A}_n=\underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^{n}f(x_i)\mathrm{\Delta }x.$

Alternatively we could have taken each sample point on the left hand side of the interval. In this case $x_i^{*}=x_{i-1}$ (see Figure 8) and the approximation becomes

$A_n=\sum _{i=1}^{n}f(x_{i-1})\mathrm{\Delta }x.$

Then the integral of f is

$\underset{a}{\overset{b}{\int }}f(x)dx=\underset{n\to \mathrm{\infty }}{\lim }{A}_n=\underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^{n}f(x_{i-1})\mathrm{\Delta }x.$

The key point is that, as long as f is continuous, these two definitions give the same answer for the integral.

Figure 7

Figure 8

Example 1

In this example we will calculate the area under the curve given by the graph of $f(x)=x$ for x between 0 and 1. First we fix an integer n and divide the interval $[0,1]$ into n subintervals of equal width. So each subinterval has width

$\mathrm{\Delta }x=\frac{1}{n}.$

To calculate the integral we will use the right-handed Riemann Sum. (We could have used the left-handed sum instead, and this would give the same answer in the end). For the right-handed sum the sample points are

$x_i^{*}=0+i\mathrm{\Delta }x=\frac{i}{n}i=1,\dots ,n$

Notice that $f(x_i^{*})=x_i^{*}=\frac{i}{n}$. Putting this into the formula for the approximation,

$A_n=\sum _{i=1}^{n}f(x_i^{*})\mathrm{\Delta }x=\sum _{i=1}^{n}f(i/n)\mathrm{\Delta }x=\sum _{i=1}^n\frac{i}{n}\cdot \frac{1}{n}=\frac{1}{n^2}\sum _{i=1}^{n}i.$

Now we use the formula

$\sum _{i=1}^{n}i=\frac{n(n+1)}{2}$

to get

$A_n=\frac{1}{n^2}\frac{n(n+1)}{2}=\frac{n(n+1)}{2n^2}.$

To calculate the integral of f between 0 and 1 we take the limit as n tends to infinity,

$\underset{0}{\overset{1}{\int }}f(x)dx=\underset{n\to \mathrm{\infty }}{\lim }\frac{n(n+1)}{2n^2}=\frac{1}{2}.$

Example 2

Next we show how to find the integral of the function $f(x)=x^2$ between x=a and x=b. This time the interval [a,b] has width b-a so

$\mathrm{\Delta }x=\frac{b-a}{n}.$

Once again we will use the right-handed Riemann Sum. So the sample points we choose are

$x_i^{*}=a+i\mathrm{\Delta }x=a+\frac{i(b-a)}{n}.$

Thus

$A_n$	$=\sum _{i=1}^{n}f(x_i^{*})\mathrm{\Delta }x$
	$=\sum _{i=1}^{n}f(a+\frac{(b-a)i}{n})\mathrm{\Delta }x$
	$=\frac{b-a}{n}\sum _{i=1}^n{(a+\frac{(b-a)i}{n})}^2$
	$=\frac{b-a}{n}\sum _{i=1}^n(a^2+\frac{2a(b-a)i}{n}+\frac{(b-a{)}^2{i}^2}{n^2})$

We have to calculate each piece on the right hand side of this equation. For the first two,

$\sum _{i=1}^n{a}^2=a^2\sum _{i=1}^{n}1=n{a}^2$

$\sum _{i=1}^n\frac{2a(b-a)i}{n}=\frac{2a(b-a)}{n}\sum _{i=1}^{n}i=\frac{2a(b-a)}{n}\cdot \frac{n(n+1)}{2}.$

For the third sum we have to use a formula

$\sum _{i=1}^n{i}^2=\frac{n(n+1)(2n+1)}{6}$

to get

$\sum _{i=1}^n\frac{(b-a{)}^2{i}^2}{n^2}=\frac{(b-a{)}^2}{n^2}\frac{n(n+1)(2n+1)}{6}.$

Putting this together

$A_n=\frac{b-a}{n}(n{a}^2+\frac{2a(b-a)}{n}\cdot \frac{n(n+1)}{2}+\frac{(b-a{)}^2}{n^2}\frac{n(n+1)(2n+1)}{6}).$

Taking the limit as n tend to infinity gives

$\underset{a}{\overset{b}{\int }}{x}^{2}dx$	$=(b-a)(a^2+a(b-a)+\frac{1}{3}(b-a{)}^2)$
	$=(b-a)(a^2+ab-a^2+\frac{1}{3}(b^2-2ab+a^2))$
	$=\frac{1}{3}(b-a)(b^2+ab+a^2)$
	$=\frac{1}{3}(b^3-a^3).$

Basic Properties of the Integral

The Constant Rule

From the definition of the integral we can deduce some basic properties. We suppose that f and g are continuous on [a,b].

Integrating Constants
If c is constant then $\underset{a}{\overset{b}{\int }}cdx=c(b-a).$

When $c>0$ and $a<b$ this integral is the area of a rectangle of height c and width b-a which equals c(b-a).

Example

$\underset{1}{\overset{3}{\int }}9dx=9(3-1)=9*2=18.$

$\underset{-2}{\overset{6}{\int }}11dx=11(6-(-2))=11*8=88.$

$\underset{2}{\overset{17}{\int }}0dx=0*(17-2)=0.$

Constant Rule
$\underset{a}{\overset{b}{\int }}cf(x)dx=c\underset{a}{\overset{b}{\int }}f(x)dx.$

When f is positive, the height of the function cf at a point x is c times the height of the function f. So the area under cf between a and b is c times the area under f. We can also give a proof using the definition of the integral, using the constant rule for limits,

$\underset{a}{\overset{b}{\int }}cf(x)dx=\underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^{n}cf(x_i^{*})=c\underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^{n}f(x_i^{*})=c\underset{a}{\overset{b}{\int }}f(x)dx.$

Example We saw in the previous section that

$\underset{0}{\overset{1}{\int }}xdx=\frac{1}{2}$.

Using the constant rule we can use this to calculate that

$\underset{0}{\overset{1}{\int }}3xdx=3\underset{0}{\overset{1}{\int }}xdx=3.\frac{1}{2}=\frac{3}{2},$

$\underset{0}{\overset{1}{\int }}-7xdx=-7\underset{0}{\overset{1}{\int }}xdx=(-7).\frac{1}{2}=-\frac{7}{2}.$

Example We saw in the previous section that

$\underset{a}{\overset{b}{\int }}{x}^{2}dx=\frac{1}{3}(b^3-a^3)$. We can use this and the constant rule to calculate that

$\underset{1}{\overset{3}{\int }}2x^{2}dx=2\underset{1}{\overset{3}{\int }}{x}^{2}dx=2.\frac{1}{3}.(3^3-1^3)=\frac{2}{3}(27-1)=\frac{52}{3}.$

The addition and subtraction rule

Addition and Subtraction Rules of Integration
$\underset{a}{\overset{b}{\int }}(f(x)+g(x))dx=\underset{a}{\overset{b}{\int }}f(x)dx+\underset{a}{\overset{b}{\int }}g(x)dx.$
$\underset{a}{\overset{b}{\int }}(f(x)-g(x))dx=\underset{a}{\overset{b}{\int }}f(x)dx-\underset{a}{\overset{b}{\int }}g(x)dx.$

As with the constant rule, the addition rule follows from the addition rule for limits:

$\underset{a}{\overset{b}{\int }}(f(x)+g(x))dx$	=	$\underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^{n}f(x_i^{*})+g(x_i^{*})$
	=	$\underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^{n}f(x_i^{*})+\underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^{n}g(x_i^{*})$
	=	$\underset{a}{\overset{b}{\int }}f(x)dx+\underset{a}{\overset{b}{\int }}g(x)dx.$

The subtraction rule can be proved in a similar way.

Example From above $\underset{1}{\overset{3}{\int }}9dx=18$ and $\underset{1}{\overset{3}{\int }}2x^{2}dx=\frac{52}{3}$ so

$\underset{1}{\overset{3}{\int }}(2x^2+9)dx=\underset{1}{\overset{3}{\int }}2x^{2}dx+\underset{1}{\overset{3}{\int }}9dx=\frac{52}{3}+18=\frac{106}{3},$

$\underset{1}{\overset{3}{\int }}(2x^2-9)dx=\underset{1}{\overset{3}{\int }}2x^{2}dx-\underset{1}{\overset{3}{\int }}9dx=\frac{52}{3}-18=-\frac{2}{3}.$

Example

$\underset{0}{\overset{2}{\int }}4x^2+14dx=4\underset{0}{\overset{2}{\int }}{x}^{2}dx+\underset{0}{\overset{2}{\int }}14dx=4\cdot \frac{1}{3}(2^3-0^3)+2\cdot 14=\frac{32}{3}+28=\frac{116}{3}.$

The Comparison Rule

Comparison Rule

• Suppose $f(x)\ge 0$ for all x in [a,b]. Then

$\underset{a}{\overset{b}{\int }}f(x)dx\ge 0.$

• Suppose $f(x)\ge g(x)$ for all x in [a,b]. Then

$\underset{a}{\overset{b}{\int }}f(x)dx\ge \underset{a}{\overset{b}{\int }}g(x)dx.$

• Suppose $M\ge f(x)\ge m$ for all x in [a,b]. Then

$M(b-a)\ge \underset{a}{\overset{b}{\int }}f(x)dx\ge m(b-a).$

If $f(x)\ge 0$ then each of the rectangles in the Riemann sum to calculate the integral of f will be above the y axis, so the area will be non-negative. If $f(x)\ge g(x)$ then $f(x)-g(x)\ge 0$ and by linearity of the integral we get the second property. Finally if $M\ge f(x)\ge m$ then the area under the graph of f will be greater than the area of rectangle with height m and less than the area of the rectangle with height M (see Figure 9). So

$M(b-a)=\underset{a}{\overset{b}{\int }}M\ge \underset{a}{\overset{b}{\int }}f(x)dx\ge \underset{a}{\overset{b}{\int }}m=m(b-a).$

Figure 9

Linearity with respect to endpoints

Additivity with respect to endpoints Suppose $a<c<b$. Then

$\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{c}{\int }}f(x)dx+\underset{c}{\overset{b}{\int }}f(x)dx.$

Again suppose that f is positive. Then this property should be interpreted as saying that the area under the graph of f between a and b is the area between a and c plus the area between c and b (see Figure 8)

Figure 8

Extension of Additivity with respect to limits of integration
When $a=b$ we have that $\mathrm{\Delta }x=\frac{b-a}{n}=0$ so

$\underset{a}{\overset{a}{\int }}f(x)dx=0.$

Also in defining the integral we assumed that a<b. But the definition makes sense even when b<a in which case $\mathrm{\Delta }x=\frac{1}{n}(b-a)$ so has changed sign. This gives

$\underset{b}{\overset{a}{\int }}f(x)dx=-\underset{a}{\overset{b}{\int }}f(x)dx.$

With these definitions,

$\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{c}{\int }}f(x)dx+\underset{c}{\overset{b}{\int }}f(x)dx$

whatever the order of a,b,c.

Fundamental Theorem of Calculus

Statement of the Fundamental Theorem

Suppose that f is continuous on [a,b]. We can define a function F by

$F(x)=\underset{a}{\overset{x}{\int }}f(t)dt\text{ for }x\text{ in }[a,b].$

Fundamental Theorem of Calculus Part I Suppose f is continuous on [a,b] and F is defined by

$F(x)=\underset{a}{\overset{x}{\int }}f(t)dt.$

Then F is differentiable on (a,b) and for all $x\in (a,b)$,

$F^{\prime }(x)=f(x).$o

Now recall that F is said to be an antiderivative of f if $F^{\prime }(x)=f(x).$.

Fundamental Theorem of Calculus Part II Suppose that f is continuous on [a,b] and that F is any antiderivative of f. Then

$\underset{a}{\overset{b}{\int }}f(x)dx=F(b)-F(a).$

Figure 1

Note: a minority of mathematicians refer to part one as two and part two as one. All mathematicians refer to what is stated here as part 2 as The Fundamental Theorem of Calculus.

Proofs

Proof of Fundamental Theorem of Calculus Part I

Suppose x is in (a,b). Pick $\mathrm{\Delta }x$ so that $x+\mathrm{\Delta }x$ is also in (a, b). Then

$F(x)=\underset{a}{\overset{x}{\int }}f(t)dt$

and

$F(x+\mathrm{\Delta }x)=\underset{a}{\overset{x+\mathrm{\Delta }x}{\int }}f(t)dt$.

Subtracting the two equations gives

$F(x+\mathrm{\Delta }x)-F(x)=\underset{a}{\overset{x+\mathrm{\Delta }x}{\int }}f(t)dt-\underset{a}{\overset{x}{\int }}f(t)dt.$

Now

$\underset{a}{\overset{x+\mathrm{\Delta }x}{\int }}f(t)dt=\underset{a}{\overset{x}{\int }}f(t)dt+\underset{x}{\overset{x+\mathrm{\Delta }x}{\int }}f(t)dt$

so rearranging this we have

$F(x+\mathrm{\Delta }x)-F(x)=\underset{x}{\overset{x+\mathrm{\Delta }x}{\int }}f(t)dt.$

According to the mean value theorem for integration, there exists a c in [x, x + Δx] such that

$\underset{x}{\overset{x+\mathrm{\Delta }x}{\int }}f(t)dt=f(c)\mathrm{\Delta }x$.

Notice that c depends on $\mathrm{\Delta }x$. Anyway what we have shown is that

$F(x+\mathrm{\Delta }x)-F(x)=f(c)\mathrm{\Delta }x$,

and dividing both sides by Δx gives

$\frac{F(x+\mathrm{\Delta }x)-F(x)}{\mathrm{\Delta }x}=f(c)$.

Take the limit as $\mathrm{\Delta }x\to 0$ we get the definition of the derivative of F at x so we have

$F^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{\lim }f(c).$.

To find the other limit, we will use the squeeze theorem. The number c is in the interval [x, x + Δx], so x≤ c ≤ x + Δx. Also, $\underset{\mathrm{\Delta }x\to 0}{\lim }x=x$ and $\underset{\mathrm{\Delta }x\to 0}{\lim }x+\mathrm{\Delta }x=x$. Therefore, according to the squeeze theorem,

$\underset{\mathrm{\Delta }x\to 0}{\lim }c=x$.

As f is continuous we have

$F^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{\lim }f(c)=f\left(\underset{\mathrm{\Delta }x\to 0}{\lim }c\right)=f(x)$

which completes the proof.

Proof of Fundamental Theorem of Calculus Part II

Define $P(x)=\underset{a}{\overset{x}{\int }}f(t)dt.$ Then by the Fundamental Theorem of Calculus part I we know that P is differentiable on (a,b) and for all $x\in (a,b)$

$P^{\prime }(x)=f(x).$

So P is an antiderivative of f. Now we were assuming that F was also an antiderivative so for all $x\in (a,b)$,

$P^{\prime }(x)=F^{\prime }(x)=f(x).$

A consequence of the Mean Value Theorem is that this implies there is a constant C such that for all $x\in (a,b)$,

$P(x)=F(x)+C.$,

and as P and F are continuous we see this holds when x=a and when x=b as well. Since we know that P(a)=0 we can put x=a into the equation to get 0=F(a) +C so C=-F(a). And putting x=b gives

$\underset{a}{\overset{b}{\int }}f(t)dx=P(b)=F(b)+C=F(b)-F(a).$

Integration of Polynomials

Using the power rule for differentiation we can find a formula for the integal of a power using the Fundamental Theorem of Calculus. Let $f(x)=x^n$. We want to find an antiderivative for f. Since the differentation rule for powers lowers the power by 1 we have that

$\frac{d}{dx}{x}^{n+1}=(n+1)x^n.$

As long as $n+1\ne 0$ we can divide by n+1 to get

$\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=x^n=f(x).$

So the function $F(x)=\frac{x^{n+1}}{n+1}$ is an antiderivative of f. If a,b>0 then F is continuous on [a,b] and we can apply the Fundamental Theorem of Calculus we can calculate the integral of f to get the following rule.

Power Rule of Integration I
$\underset{a}{\overset{b}{\int }}{x}^{n}dx=[\frac{x^{n+1}}{n+1}{]}_a^b=\frac{b^{n+1}-a^{n+1}}{n+1}.$ as long as $n\ne -1$ and $a,b>0.$

Notice that we allow all values of n, even negative or fractional. If n>0 then this works even if a or b are negative.

Power Rule of Integration II
$\underset{a}{\overset{b}{\int }}{x}^{n}dx=[\frac{x^{n+1}}{n+1}{]}_a^b=\frac{b^{n+1}-a^{n+1}}{n+1}.$ as long as $n>0.$

Example

To find $\underset{1}{\overset{2}{\int }}{x}^{3}dx$ we raise the power by 1 and have to divide by 4. So $\underset{1}{\overset{2}{\int }}{x}^{3}dx=[\frac{x^4}{4}{]}_1^2=\frac{2^4}{4}-\frac{1^4}{4}=\frac{15}{4}.$

Example

The power rule also works for negative powers. For instance $\underset{1}{\overset{3}{\int }}\frac{1}{x^3}dx=\underset{1}{\overset{3}{\int }}{x}^{-3}dx=[\frac{x^{-2}}{-2}{]}_1^3=\frac{1}{-2}(3^{-2}-1^{-2})=-\frac{1}{2}(\frac{1}{3^2}-1)=-\frac{1}{2}(\frac{1}{9}-1)=\frac{1}{2}\cdot \frac{8}{9}=\frac{4}{9}.$

Example

We can also use the power rule for fractional powers. For instance $\underset{0}{\overset{5}{\int }}\sqrt{x}dx=\underset{0}{\overset{5}{\int }}{x}^{\frac{1}{2}}dx=[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}dx{]}_0^5=\frac{2}{3}(5^{\frac{3}{2}}-0^{\frac{3}{2}})=\frac{2}{3}\left(5^{\frac{3}{2}}\right)$

Example

Using the linearity rule we can now integrate any polynomial. For example

$\underset{0}{\overset{3}{\int }}(3x^2+4x+2)dx=[x^3+2x^2+2x{]}_0^3=3^3+2\cdot 3^2+2\cdot 3-0=27+18+6=51.$

Indefinite Integrals

The Fundamental Theorem of Calculus tells us that if f is continuous then the function $F(x)=\underset{a}{\overset{x}{\int }}f(t)dt$ is an antiderivative of f (i.e. $F^{\prime }(x)=f(x)$. However it is not the only antiderivative. We can add any constant to F without changing the derivative.

We write $F(x)=\int f(x)dx+C$ if the derivative of F is $F^{\prime }(x)=f(x).$

Example

Since the derivative of $x^4$ is $4x^3$ the general antiderivative of $4x^3$ is $x^4$ plus a constant. Thus

$\int 4x^{3}dx=x^4+C.$

Example: Finding antiderivatives

Let us return to the previous example, that of $6x^2$. How would we go about finding the integral of this function? Recall the rule from differentiation that

$\frac{d}{dx}{x}^n=n{x}^{n-1}$

In our circumstance, we have:

$\frac{d}{dx}{x}^3=3x^2$

This is a start! We now know that the function we seek will have a power of 3 in it. How would we get the constant of 6? Well,

$2\frac{d}{dx}{x}^3=2\times 3x^2=6x^2.$

Thus, we say that $2x^3$ is an antiderivative of $6x^2$.

Basic Properties of Indefinite Integrals

Constant Rule for indefinite integrals
If c is constant then $\int cf(x)dx=c\int f(x)dx.$

Sum/Difference Rule for indefinite integrals

$\int f(x)+g(x)dx=\int f(x)dx+\int g(x)dx.$

$\int f(x)-g(x)dx=\int f(x)dx-\int g(x)dx.$

Indefinite integrals of Polynomials

Since

$\frac{d}{dx}\frac{1}{n+1}{x}^{n+1}=x^n$

we have the following rule for indefinite integrals.

Power rule for indefinite integrals If $n\ne -1,$ then

$\int x^{n}dx=\frac{1}{n+1}{x}^{n+1}+C.$

Integral of the Inverse function

Since

$\frac{d}{dx}\ln x=\frac{1}{x}$

We know that

$\int \frac{dx}{x}=\ln \left|x\right|+C$

Note that the polynomial integration rule does not apply when the exponent is -1. This technique of integration must be used instead. Since the argument of the natural logarithm function must be positive (on the real line), the absolute value signs are added around its argument to ensure that the argument is positive.

Integral of Sine and Cosine

In this section we will concern ourselves with determining the integrals of the sin and cosine function.

Recall that

$\frac{d}{dx}\sin x=\cos x$

$\frac{d}{dx}\cos x=-\sin x.$

So sin x is an antiderivative of cos x and -cos x is an antiderivative of sin x. Hence we get the following rules for integrating sin x and cos x.

$\int \cos xdx=\sin x+C$

$\int \sin xdx=-\cos x+C$

We will find how to integrate more complicated trigonometric functions in the chapter on Further integration techniques.

Integral of the Exponential function

Since

$\frac{d}{dx}{e}^x=e^x$

we see that $e^x$ is its own antiderivative. Perhaps a more useful definition of this rule can be given as:

$\frac{d}{dx}{e}^{f(x)}=f^{\prime }(x)\cdot e^{f(x)}$

hence:

$\frac{d}{dx}{e}^x=(1)e^x$

Where the exponent (x) is differentiated to give a value of 1

Simplified: $\frac{d}{dx}{e}^x=(1)e^x$

Becomes: $\frac{d}{dx}{e}^x=e^x$

So the integral of an exponential function can be found thusly:

$\int e^{x}dx=e^x+C$

Integration Rules

$\int ndx=nx+C$

$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C,n\ne -1$

$\int e^{x}dx=e^x+C$

$\int a^{x}dx=\frac{a^x}{\ln \left(a\right)}+C$

$\int \frac{1}{x}dx=\ln \left|x\right|+C,x>0$

$\int \sin xdx=-\cos x+C$

$\int \cos xdx=\sin x+C$

$\int \tan xdx=-\ln |\cos x|+C$

$\int \csc xdx=-\ln |\csc x+\cot x|+C$

$\int \sec xdx=\ln |\sec x+\tan x|+C$

$\int \cot xdx=\ln |\sin x|+C$

$\int {\sec }^{2}xdx=\tan x+C$

$\int {\csc }^{2}xdx=-\cot x+C$

$\int \sec x\tan xdx=\sec x+C$

$\int \csc x\cot xdx=-\csc x+C$

$\int \frac{1}{\sqrt{a^2-b^2{x}^2}}dx=\frac{1}{b}\arcsin \frac{bx}{a}+C$

$\int \frac{1}{a^2+b^2{x}^2}dx=\frac{1}{ab}\arctan \frac{bx}{a}+C$

$\int \frac{1}{x\sqrt{x^2-a^2}}dx=\frac{1}{a}\mathrm{arcsec}\frac{\left|x\right|}{a}+C$

The Substitution Rule

Suppose that we want to find $\int (\cos (x^2)\cdot x)dx.$

The Fundamental theorem of calculus tells us that we want to find an antiderivative of the function:

$f(x)=\cos (x^2)\cdot x$

Since $\sin (x)$ differentiates to $\cos (x)$ as a first guess we might try the function $\sin (x^2)$. But by the Chain Rule

$\frac{d}{dx}\sin (x^2)=\cos (x^2)\cdot \frac{d}{dx}{x}^2=\cos (x^2)\cdot 2x=2x\cos (x^2)$

which is almost what we wanted apart from the fact that there is an extra factor of 2 in front. But this is easily dealt with because we can divide by any constant so

$\frac{d}{dx}\frac{\sin (x^2)}{2}=\frac{1}{2}\cdot \frac{d}{dx}\sin (x^2)=\frac{1}{2}\cdot 2\cos (x^2)x=x\cos (x^2)=f(x).$

So using the Fundamental Theorem of Calculus, $\int x\cos (x^2)dx=\frac{\sin (x^2)}{2}+C.$

In fact this technique will work for more general integrands. Suppose u is a differentiable function. Then to evaluate $\int u^{\prime }(x)cos(u(x))dx$ we just have to notice that by the Chain Rule

$\frac{d}{dx}\sin (u(x))=\cos (u(x))\frac{du}{dx}=u^{\prime }(x)\cos (u(x)).$

As long at $u^{\prime }$ is continuous the Fundamental Theorem applies and tells us that

$\int \cos (u(x))u^{\prime }(x)dx=\sin (u(x))+C.$

Now the right hand side of this equation is just the intgral of $\cos (u)$ but with respect to u. If we write u instead of u(x) this becomes $\int \cos (u(x))u^{\prime }(x)dx=\sin (u)du+C=\int \cos (u)du.$

So for instance if $u(x)=x^3$ we have worked out that

$\int (\cos (x^3)\cdot 3x^2)dx=\sin (x^3)+C.$

Now there was nothing special about using the cosine function in the discussion above, and it could be replaced by any other function. Doing this gives us the substitution rule for indefinite integrals.

Substitution rule for indefinite integrals Assume u is differentiable with continuous derivative and that f is continuous on the range of u. Then $\int f(u(x))\frac{du}{dx}dx=\int f(u)du.$

Notice that it looks like you can cancel in the expression $\frac{du}{dx}dx$ to leave just a $du$. This does not really make any sense as $\frac{du}{dx}$ is not a fraction, but is a good way to remember the substitution rule.

There is a similar rule for definite integrals, but we have to change the endpoints.

Substitution rule for definite integrals Assume u is differentiable with continuous derivative and that f is continuous on the range of u. Suppose $c=u(a)$ and $d=u(b)$. Then $\underset{a}{\overset{b}{\int }}f(u(x))\frac{du}{dx}dx=\underset{c}{\overset{d}{\int }}f(u)du.$

Examples

Consider the integral

$\underset{0}{\overset{2}{\int }}x\cos (x^2+1)dx$

By using the substitution u = x² + 1, we obtain du = 2x dx and

$\underset{0}{\overset{2}{\int }}x\cos (x^2+1)dx$	$=\frac{1}{2}\underset{0}{\overset{2}{\int }}\cos (x^2+1)2xdx$
	$=\frac{1}{2}\underset{1}{\overset{5}{\int }}\cos (u)du$
	$=\frac{1}{2}(\sin (5)-\sin (1)).$

Note how the lower limit x = 0 was transformed into u = 0² + 1 = 1 and the upper limit x = 2 into u = 2² + 1 = 5.

Proof of the substitution rule

We will now prove the substitution rule for definite integrals. Let F be an anti derivative of f so

$F^{\prime }(x)=f(x)$. By the Fundamental Theorem of Calculus

$\underset{c}{\overset{d}{\int }}f(u)du=F(d)-F(c).$

Next we define a function $G$ by the rule

$G(x)=F(u(x)).$

Then by the Chain rule G is differentiable with derivative

$G^{\prime }(x)=F^{\prime }(u(x))u^{\prime }(x)=f(u(x))u^{\prime }(x).$

Integrating both sides with respect to x and using the Fundamental Theorem of Calculus we get

$\underset{a}{\overset{b}{\int }}f(u(x))u^{\prime }(x)dx=\underset{a}{\overset{b}{\int }}{G}^{\prime }(x)dx=G(b)-G(a).$

But by the definition of F this equals

$G(b)-G(a)=F(u(b))-F(u(a))=F(d)-F(c)=\underset{c}{\overset{d}{\int }}f(u)du.$

Hence

$\underset{a}{\overset{b}{\int }}f(u(x))u^{\prime }(x)dx=\underset{c}{\overset{d}{\int }}f(u)du.$

which is the substitution rule for definite integrals.

Integration of even and odd functions

Recall that a function f is called odd if it satisfies $f(-x)=-f(x)$ and is called even if $f(-x)=f(x).$

Suppose f is a continuous odd function then for any a,

$\underset{-a}{\overset{a}{\int }}f(x)dx=0.$

If f is a continuous even function then for any a,

$\underset{-a}{\overset{a}{\int }}f(x)dx=2\underset{0}{\overset{a}{\int }}f(x)dx.$

Caution: For improper integrals (e.g. if a is infinity, or if the function approaches infinity at 0 or a, etc.), the first equation above is only true if $\underset{0}{\overset{a}{\int }}f(x)dx$ exists. Otherwise the integral is undefined, and only the Cauchy principal value is 0.

Suppose f is an odd function and consider first just the integral from -a to 0. We make the substitution u=-x so du=-dx. Notice that if x=-a then u=a and if x=0 then u=0. Hence $\underset{-a}{\overset{0}{\int }}f(x)dx=-\underset{a}{\overset{0}{\int }}f(-u)du=\underset{0}{\overset{a}{\int }}f(-u)du.$ Now as f is odd, $f(-u)=-f(u)$ so the integral becomes $\underset{-a}{\overset{0}{\int }}f(x)dx=-\underset{0}{\overset{a}{\int }}f(u)du.$ Now we can replace the dummy variable u with any other variable. So we can replace it with the letter x to give $\underset{-a}{\overset{0}{\int }}f(x)dx=-\underset{0}{\overset{a}{\int }}f(u)du=-\underset{0}{\overset{a}{\int }}f(x)dx.$

Now we split the integral into two pieces

$\underset{-a}{\overset{a}{\int }}f(x)dx=\underset{-a}{\overset{0}{\int }}f(x)dx+\underset{0}{\overset{a}{\int }}f(x)dx=-\underset{0}{\overset{a}{\int }}f(x)dx+\underset{0}{\overset{a}{\int }}f(x)dx=0.$

The proof of the formula for even functions is similar, and is left as an exercise.

Integration by Parts

Integration by parts for indefinite integrals Suppose f and g are differentiable and their derivatives are continuous. Then

$\int f(x)g^{\prime }(x)dx=f(x)g(x)-\int f^{\prime }(x)g(x)dx.$

If we write u=f(x) and v=g(x) then using the Leibnitz notation du=f'(x) dx and dv=g'(x) dx and the integration by parts rule becomes

$\int udv=uv-\int vdu.$

For definite integrals the rule is essentially the same, as long as we keep the endpoints.

Integration by parts for definite integrals Suppose f and g are differentiable and their derivatives are continuous. Then

$\underset{a}{\overset{b}{\int }}f(x)g^{\prime }(x)dx=[f(x)g(x){]}_a^b-\underset{a}{\overset{b}{\int }}{f}^{\prime }(x)g(x)dx$

$=f(b)g(b)-f(a)g(a)-\underset{a}{\overset{b}{\int }}{f}^{\prime }(x)g(x)dx$.

This can also be expressed in Leibniz notation.

$\underset{a}{\overset{b}{\int }}udv=[uv{]}_a^b-\underset{a}{\overset{b}{\int }}vdu.$

Example Find

$\int x\cos (x)dx$

Here we let:

$u=x$, so that $du=dx$,

$dv=\cos (x)dx$ , so that $v=\sin (x)$.

Then:

$\int x\cos (x)dx$	$=\int udv$
	$=uv-\int vdu$

$\int x\cos (x)dx=x\sin (x)-\int \sin (x)dx$

$\int x\cos (x)dx=x\sin (x)+\cos (x)+C$

where C is an arbitrary constant of integration.

Example

$\int x^2{e}^{x}dx$

In this example we will have to use integration by parts twice.

Here we let

$u=x^2$, so that $du=2xdx$,

$dv=e^{x}dx$, so that $v=e^x$.

Then:

$\int x^2{e}^{x}dx$	$=\int udv$
	$=uv-\int vdu$

$\int x^2{e}^{x}dx=x^2{e}^x-\int 2x{e}^{x}dx=x^2{e}^x-2\int x{e}^{x}dx.$

Now to calculate the last integral we use integration by parts again. Let

$u{=}^{″}x$, so that $du=dx$,

$dv=e^{x}dx$, so that $v=e^x$

and integrating by parts gives

$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x.$

So in the end

$\int x^2{e}^{x}dx=x^2{e}^x-2(x{e}^x-e^x)=x^2{e}^x-2x{e}^x+2e^x=e^x(x^2-2x+2).$

Example Find

$\int \ln (x)dx.$

The trick here is to write this integral as

$\int \ln (x)\cdot 1dx.$

Now let

$u=ln(x)$ so $du=1/xdx$,

$v=x$ so $dv=1dx$.

Then using integration by parts,

$\int \ln (x)dx$	$=x\ln (x)-\int \frac{x}{x}dx$
	$=x\ln (x)-\int 1dx$

$\int \ln (x)dx=x\ln (x)-x+C$

$\int \ln (x)dx=x(\ln (x)-1)+C$

where, again, C is an arbitrary constant.

Example Find $\int \arctan (x)dx.$

Again the trick here is to write the integrand as $\arctan (x)=\arctan (x)\cdot 1$. Then let

u = arctan(x); du = 1/(1+x²) dx

v = x; dv = 1·dx

so using integration by parts,

$\int \arctan (x)dx$	$=x\arctan (x)-\int \frac{x}{1+x^2}dx$
	$=x\arctan (x)-\frac{1}{2}\ln (1+x^2)+C.$

Example Find :$\int e^x\cos (x)dx$ This example uses integration by parts twice. First let,

u = e^x; thus du = e^xdx

dv = cos(x)dx; thus v = sin(x)

so

$\int e^x\cos (x)dx=e^x\sin (x)-\int e^x\sin (x)dx$

Now, to evaluate the remaining integral, we use integration by parts again, with

u = e^x; du = e^xdx

v = -cos(x); dv = sin(x)dx

Then

$\int e^x\sin (x)dx$	$=-e^x\cos (x)-\int -e^x\cos (x)dx$
	$=-e^x\cos (x)+\int e^x\cos (x)dx$

Putting these together, we get

$\int e^x\cos (x)dx=e^x\sin (x)+e^x\cos (x)-\int e^x\cos (x)dx$

Notice that the same integral shows up on both sides of this equation. So we can simply add the integral to both sides to get:

$2\int e^x\cos (x)dx=e^x(\sin (x)+\cos (x))$

$\int e^x\cos (x)dx=\frac{e^x(\sin (x)+\cos (x))}{2}$