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.
Suppose that there exists a function g, of another variable , such that and let, . So that,
Now, by the chain rule,
The gradient of a smooth curve, , at a point is the gradient of the tangent to the curve at the point . Point is on the curve and is a neighbouring point whose value is displaced a small quantity, .
The idea behind differentiation is that as becomes very small, the gradient of tends towards the gradient of the curve. In the limit as becomes infinitesimally close to zero, the gradient becomes the gradient of the curve.
there is a fair bit of analytic work missing (higher education) to make these ideas sound.
We also write:
Standard results can be proved for different functions.
If , then we need to consider the small angle approximation that is if radians is very small (infinitesimal), then and , 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, where ,
Integration is introduced as the reversal of differentiation i.e. in solving a differential equation, . 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, , defined by the area under between and and a general point, . If a small increment, , is applied to giving a small element, of area. Now,
dividing though by , gives,
a limit sandwich where, as ,
The curve function, is the derivative of the area function; hence the area function is the anti-derivative of the curve function and,