Table of Contents
Why is the integral of a function the area?
You can divide up the area between x=A and x=B under a function by putting a mess of rectangles under it. The area under the function (the integral) is given by the antiderivative! Again, this approximation becomes an equality as the number of rectangles becomes infinite.
What does it mean to integrate over the x-axis?
A general function f(x) is sometimes positive and sometimes negative, so the integral calculates the signed area, that is, the total area above the x-axis minus the total area below the x-axis. Detailed description of diagram. The integral counts areas above/below the x-axis as positive/negative.
What does the integral represent graphically?
A definite integral of a function can be represented as the signed area of the region bounded by its graph.
What is the definite integral of f(x) from a to B?
Given a function f (x) f ( x) that is continuous on the interval [a,b] [ a, b] we divide the interval into n n subintervals of equal width, Δx Δ x, and from each interval choose a point, x∗ i x i ∗. Then the definite integral of f (x) f ( x) from a a to b b is
What does f(x)dx mean?
f(x)dx is called the definite integral of f(x) over the interval [a,b] and stands for the area underneath the curve y = f(x) over the interval [a,b] (with the understanding that areas above the x-axis are considered positive and the areas beneath the axis are considered negative).
What is the area under the function given by the integrals?
The area under the function (the integral) is given by the antiderivative! Again, this approximation becomes an equality as the number of rectangles becomes infinite. As an aside (for those of you who really wanted to read an entire post about integrals), integrals are surprisingly robust.
What is the area under the function given by the antiderivative?
Holy crap! The area under the function (the integral) is given by the antiderivative! Again, this approximation becomes an equality as the number of rectangles becomes infinite. As an aside (for those of you who really wanted to read an entire post about integrals), integrals are surprisingly robust.