Question #22e59

1 Answer
Sep 12, 2015

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 xx values into your current function. If F(x)F(x) was the antiderivative and f(x)f(x) was the function:

F(x) = int f(x)dx = int x^2dx = x^3/3 + CF(x)=f(x)dx=x2dx=x33+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.


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.

![https://upload.wikimedia.org/](useruploads.socratic.org)

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 Nth rectangle that gives you the exact area, you're done. The definite integral gives us this result.