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 #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#.


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.

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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.