## Integration by Substitution

Current UK exam textbooks pass over proofs and mathematical discussions in a hurry to show the ‘how to’ of exam questions.

Integration by substitution is a little more than just backwards chain-rule and deserves a fuller treatment.

Try this,

Let,

$y=\displaystyle \int \textnormal{f}(x) \ \textnormal{d}x$

then,

$\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{f}(x)$

Suppose that there exists a function g, of another variable $u$, such that $x=\textnormal{g}(u)$ and let, $\textnormal{f}(x)=\textnormal{f}(\textnormal{g}(u))=\textnormal{F}(u)$. So that,

$\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{F}(u)$

Now, by the chain rule,

$\dfrac{\textnormal{d}y}{\textnormal{d}u}=\dfrac{\textnormal{d}y}{\textnormal{d}x}\times \dfrac{\textnormal{d}x}{\textnormal{d}u}=\textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u}$

Hence,

$y=\displaystyle \int \dfrac{\textnormal{d}y}{\textnormal{d}u} \ \textnormal{d}u=\displaystyle \int \textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u} \ \textnormal{d}u$

i.e.

$y=\displaystyle \int \textnormal{f}(x) \ \textnormal{d}x=\displaystyle \int \textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u} \ \textnormal{d}u$

## Differentiation From First Principles

The gradient of a smooth curve, $\textnormal{f}(x)$, at a point $x$ is the gradient of the tangent to the curve at the point $x$. Point $P$ is on the curve and $Q$ is a neighbouring point whose $x$ value is displaced a small quantity, $\delta x$.

The idea behind differentiation is that as $\delta x$ becomes very small, the gradient of $PQ$ tends towards the gradient of the curve. In the limit as $\delta x$ becomes infinitesimally close to zero, the gradient $PQ$ becomes the gradient of the curve.

We write:

$\textnormal{gradient f}(x)=\dfrac{\textnormal{d}y}{\textnormal{d}x}=\lim_{\delta x \rightarrow 0}\left(\dfrac{\delta y}{\delta x}\right)=\lim_{\delta x \rightarrow 0}\left(\dfrac{\textnormal{f}(x+\delta x)-\textnormal{f}(x)}{\delta x}\right)$

there is a fair bit of analytic work missing (higher education) to make these ideas sound.

We also write:

$\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{f}'(x).$

STANDARD RESULTS

Standard results can be proved for different functions.

If $\textnormal{f}(x)=x^{n}$ then

If $\textnormal{f}(x)=\sin x$, then we need to consider the small angle approximation that is if $\delta x$ radians is very small (infinitesimal), then $\delta x\approx\sin \delta x$ and $\cos \delta x \approx 1$, and compound trigonometry from which follows,

The differentiation process described above is linear and extends to more complicated functions. That is to say that if, $y=a\textnormal{f}(x)+b\textnormal{g}(x)$ where $a,b \in \mathbb{R}$,
$\dfrac{\textnormal{d}y}{\textnormal{d}x}=a\textnormal{f}'(x)+b\textnormal{g}'(x)$

## The Fundamental Theorem of Calculus

Integration is introduced as the reversal of differentiation i.e. in solving a differential equation, $\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{g}(x)$. The link between integration and area is often passed over and is the subject of the Fundamental Theorem of Calculus. [The following discussion can be adapted for a decreasing function or, piece-wise, a function which successively increases or decreases.]

Consider and area function, $A(x)$, defined by the area under $\textnormal{f}(x)$ between $a$ and and a general point, $x$. If a small increment, $\delta x$, is applied to $x$ giving a small element, $\delta A$ of area. Now,

$\textnormal{f}(x)\delta x \leqslant \delta A \leqslant \textnormal{f}(x+\delta x)\delta x$

dividing though by $\delta x$, gives,

$\textnormal{f}(x) \leqslant \dfrac{\delta A}{\delta x} \leqslant \textnormal{f}(x+\delta x),$

a limit sandwich where, as $\delta x \rightarrow 0$,

$\dfrac{\textnormal{d}A}{\textnormal{d}x}=\textnormal{f}(x)$

The curve function, $\textnormal{f}(x)$ is the derivative of the area function; hence the area function is the anti-derivative of the curve function and,

$\displaystyle\int \textnormal{f}(x) \textnormal{d}x=A(x).$