Question

Question #22e59

Answer 1

An indefinite integral tells you the antiderivative of a function. That is, it tells you what the original function was whose slope can be acquired by plugging `x` values into your current function. If `F(x)` was the antiderivative and `f(x)` was the function:

`F(x) = int f(x)dx = int x^2dx = x^3/3 + C`

`d/(dx)[F(x)] = d/(dx)[x^3/3 + C] = cancel(d/(dx))[cancel(int) f(x)dx] = f(x) = x^2`

The `+C` tells you that vertical shifts in `F(x)` still result in the same `f(x)` since the derivative of a constant is `0`.

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A basic definite integral often is for acquiring the area between a curve and an axis. It is an indefinite integral with specific boundaries in which you would evaluate the area. So, if you wanted the area under `x^2` from `x = 0` to `x = 2`:

graph{(y-x^2)(y)sqrt(1^2 - (x-1)^2)/sqrt(1^2 - (x-1)^2) <= 0 [-6.64, 7.407, -2.246, 4.777]}

then you would solve:

`int_0^2 x^2dx`

The formal definition of the definite integral is:

`int_a^bf(x)dx = lim_(N->oo) sum_(i = 1)^N f(x_i^"*")Deltax_i`

All this really means is that you are taking a large number of rectangles of varying heights `f(x)` with constant width `Deltax` and increasing the number of rectangles you use until you have covered the entire area under the curve using a bunch of rectangles.

This is therefore just an approximation to the area, and then making the approximation better and better. The more rectangles you use, the thinner `Deltax` is since the integration interval `(a->b)` never changes, and the better the approximation you get.

When you get to the `N`th rectangle that gives you the exact area, you're done. The definite integral gives us this result.