Integration

Excerpts

• Integration

  The remainder needed will always be equal to the last term added.

• Integration

  A microscope would not show even the $18^{\text{th}}$ term! So the infinite number of operations is no such dreadful thing after all.

• Integration

  Clearly, at any point $P$ of the curve, the value of $y$ will be the sum of all the little $dy$’s from $0$ up to that level, that is to say, $\ds\int dy = y$.

• Integration

  But, as in the previous case, this requires the addition of an undetermined constant $C$, because we have not been told at what height above the origin the curve will begin, when $x = 0$.

• Integration

  Any one can understand how the whole of anything can be conceived of as made up of a lot of little bits; and the smaller the bits the more of them there will be.

• Integration

  But $x$ began by being $0$, and increases to the particular value of $x$ at the point $P$, so that its average value from $0$ to that point is $\frac{1}{2}x$. Hence $\ds\int \tfrac{1}{5} x\, dx = \tfrac{1}{10} x^2$; or $y=\frac{1}{10}x^2$.

• Integration

  The simple reason is that there are a vast number of cases in which one cannot calculate the bigness of the thing as a whole without reckoning up the sum of a lot of small parts. The process of “*integrating*” is to enable us to calculate totals that otherwise we should be unable to estimate directly.

• Integration

  If at any point of the operation we stop, there will still be a piece wanting to make up the whole $2$ inches; and the piece wanting will always be the same size as the last piece added.

• Integration

  If we want to go so far that not even a Whitworth’s measuring machine would detect it, we should merely have to go to about $20$ terms. A microscope would not show even the $18^{\text{th}}$ term! So the infinite number of operations is no such dreadful thing after all.

• Integration

  For we have seen that differentiating a curve means finding an expression for its slope (or for its slopes at different points). Can we perform the reverse process of reconstructing the whole curve if the slope (or slopes) are prescribed for us?

• Integration

  As the only information we have is as to the slope, we are without any instructions as to the particular height above $O$; in fact the initial height is undetermined. The slope will be the same, whatever the initial height.

Equations

• Integration

  \ds\int dy = y

  The integral sign summed over the small pieces dy gives the total y, so integrating dy recovers y.

• Integration

  \ds\int dx = x

  The integral of the small pieces dx gives the total x.

• This equation is in Geometrical Meaning of Differentiation (Geometrical Meaning of Differentiation)
• This equation is in Geometrical Meaning of Differentiation (Geometrical Meaning of Differentiation)

• Integration

  \dfrac{y}{x} = a

  If y and x are the totals of all the dy's and dx's of a constant-slope line, their ratio equals the slope a.

• Integration

  y = ax + C.

  Reconstructing a line from its constant slope gives y = ax + C, where the undetermined constant C is the height at x = 0.

• Integration

  \frac{dy}{dx} = ax.

  The slope of the curve increases in proportion to x.

• Integration

  \frac{dy}{dx} = \tfrac{1}{5} x

  A concrete case of the increasing slope, with a = 1/5, so the slope equals x/5.

• Integration

  \ds\int \tfrac{1}{5} x\, dx = \tfrac{1}{10} x^2

  The sum of the pieces (1/5)x dx equals (1/10)x^2, found by using the average value of x, which is x/2, from 0 to x.

• Integration

  y=\frac{1}{10}x^2

  The height y of the curve with slope x/5 is x^2/10, up to an added constant.

• Integration

  y = \tfrac{1}{10}x^2 + C.

  The equation of the curve with slope x/5 includes the undetermined constant C, because the starting height above the origin was not given.

Problems