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Some confusion can arise because it can either give you a number (which is the area under the curve) but you can also use it just to get an equation (the opposite to differentiation).

Let's look at the symbol more closely and break it down.

∫ | b | f(x) dx |

a |

The upper and lower limits show that you just want the area under a curve from the lower limit to the upper one.

Now the dx part is intended to suggest dividing the area under the curve into an infinite number of rectangles with an infinitesimally small width (shown below) and not a number that is multiplied by something else although it is sometimes treated as if it is a number you can manipulate e.g.

∫ |
dx x |
= | ∫ | 1 x | dx |

Now I shall explain definite integration...

Let's draw some function, y = something:

Now the Δx just means a difference between a higher x value and a lower x value which we can use for the width of each rectangle. The area of the rectangle is the y value times the Δx value (y•Δx, which is the same as width times height). The green area means the part which we have not calculated. This makes our answer an approximation. One way to get a more accurate answer is to reduce the width of the rectangles and use more of them although you would not use this method (it is impractical to use a million rectangles, for example), you could use the Simpson's rule or another method. I am using this method just to define what integration is.

By making Δx smaller and smaller (which is written Δx → 0, as Δx approches 0), we get a definition of the integral sign. Don't worry if you do not understand these symbols, I am just presenting them for those that do:

lim | ∑ | y_{i}•Δx |
This just means a sum all the y values times Δx (a sum of all the areas of the rectangles) |

Δx → 0 |

= | ∫ | b | y dx Now this is not an approximation. It is like the above but with unimaginally small rectangles |

a |

After you have found an equation, you use upper and lower boundaries but what are we doing when we do something like this:

[2x] | 10 2 |

By putting a number in the equation, you can imagine that it is like the area from 0 up to that point. If I subtract the smaller area from the larger one, I am left with the area between the upper and lower boundary. Here is what I mean:

By putting in a smaller number in to the equation, it corresponds to an area up to that point:

By putting in a larger number, it corresponds to an area up to that one too:

So if I subtract these areas, I get the one in between:

minus | equals |

Note that you can have negative boundaries, or even swap them around and get a negative result.

Also people usually ignore the constant you get from integrating because you are effectively subtracting an unknown constant from itself, leaving 0. Let's look at an example with the unknown constant left in:

∫ 2 dt = 2t + C this is what I will use for a definite integral:

∫ | 10 | 2 dt |

2 |

= [2t + C] | 10 2 |

so you would calculate this like (2 ×10 + C) - (2 ×2 + C) = (20 + C) - (4 + C) = 20 -4 + C - C = 16. Since this unknown constant always disappears in this way, it is ignored.

Another point is that when you use certain methods to integrate like trigonometric substitution or functions of a linear function of x, you change the equation AND the dx bit. The reason you have to change both is because when you change the equation, that is like changing the equation representing the height of the rectangles so you also need to change the width of them so you change the dx part too.

Now on to indefinite integration...

d dx |
2x = 2 then ∫ 2 dx = 2x + C |

The constant of integration

When you differentiate a number, it is always zero (the gradient or 'steepness' of a horizontal line is zero) so this information is lost.

Let's look at a line y = x

d dx |
x | 2 | = 2x but also | d dx | x | 2 | + 5 = 2x |

∫ 2x dx = x

This can be found if you are given initial conditions for example, if you are trying to calculate the voltage in a circuit, you may get something like:

- | 1 R |
ln(E-Ri) = | t L | + C |

- | 1 R |
ln(E-R0) = | 0 L | + C |

- | 1 R |
ln(E) = | C |

Have you found an error or do you want to add more
information to these pages? You can contact me at the bottom of the home page. |