Integration

Integration is the Inverse of Differentiation

Integration is the inverse of differentiation. This is the Fundamental Theorem of Calculus.

Example: We know that the derivative of $x^{2}$ is $2x$. This means that the integral of $2x$ is $x^{2}$, which we write as $int2xdx=x^{2}$. Do not forget the $dx$, it will be important in later sections.

However, this definition of integration causes a problem.

Example: Differentiate

i) $x^{3}$

ii) $x^{3}+3$

iii) $x^{3}-19$

You will notice that all of these are $3x^{2}$. So what is $int3x^{2}dx$? Is it $x^{3}$, $x^{3}+3$ or $x^{3}-19$? Under the rule established above it is all three.

This is where the constant of integration comes in.

In fact, we write $int3x^{2}dx=x^{3}+c$, where $c$ represents any constant, because all constants differentiate to $0$. We do this for all indefinite integrals.

Integrating a Polynomial

To integrate something of the form $x^{n}$, add $mathbb{1}$ to the power then divide by the new power, to get $dfrac{1}{n+1}x^{n+1}$

$int x^{n}dx=dfrac{1}{n+1}x^{n+1}+c$

As with differentiation, we can integrate term by term, and the integral of $n$ times a function is $n$ times the integral of a function.

This means that we can now integrate polynomials.

Example: Integrate $x^{2}+3x+4$

$int x^{2}+3x+4dx=int x^{2}dx+3int xdx+4int1dx$

Do the integration term by term.

For $x^{2}$, we add $1$ to the power to get $3$, then divide by the new power, which is $3$, to get $dfrac{1}{3}x^{3}$

For $x=x^{1}$, we add $1$ to the power to get $2$, then divide by the new power, which is $2$, to get $dfrac{1}{2}x^{2}$

For $1=x^{0}$, we add $1$ to the power to get $1$, then divide by the power, which is $1$, to get $x$.

Putting it all together, and not forgetting $+c$:

$begin{aligned}int x^{2}+3x+4dx&=dfrac{1}{3}x^{3}+left( 3timesdfrac{1}{2}x^{2}right) +left( 4times xright) +c[1.2em]&=dfrac{1}{3}x^{3}+dfrac{3}{2}x^{2}+4x+cend{aligned}$

Note: Our rule works for anything of the form $x^{n}$, not just positive whole numbers, so we can integrate far more than just polynomials. Also, it does not work for $n=-1$ because this would mean dividing by $0$.

Finding the Constant of Integration

Sometimes you will be told that an integral passes through some value, and based on this asked to find the constant of integration.

Example: The curve $y=f(x)$ passes through the point $(8,200)$ and $f'(x)=9x-5$. Find $f(x)$.

$f'(x)=9x-5$

Integrate both sides,

$begin{aligned}f(x)&=intleft( 9x-5right) dx[1.2em]&=left( 9timesint xdxright) -left( 5timesint1dxright) [1.2em]&=left( 9timesdfrac{1}{2}x^{2}right) -5x+c[1.2em]&=dfrac{9}{2}x^{2}-5x+cend{aligned}$

$f(8)=200$

$left( dfrac{9}{2}times8^{2}right) -left( 5times8right) +c=200$

$left( dfrac{9}{2}times64right) -40+c=200$

$288-40+c=200$

$248+c=200$

$c=-48$

$f(x)=dfrac{9}{2}x^{2}-5x-48$