The integral of a function $f(x)$, $∫_{0}^{b}f(x)dx $, gives the area enclosed between the curve and the $x$-axis, between the $y$-axis and $x=b$.

Similarly, the integral $∫_{a}^{b}f(x)dx$, returns the area between the curve and the $x$-axis, between $x=a$ and $x=b$.

If $f$ is continuous in $[a,b]$ and if $F$ is any anti-derivative of $f$ on $[a,b]$, then $∫_a^{b}f(x)dx=F(b) - F(a)$

This formulation was set down by Augustin-Lewis Cauchy, 1789-1857, a French mathematician, and in so doing he finally and formally united the two branches of calculus, that of Newton, and that of Leibniz.

If $y_1$ and $y_2$ are continuous on $a ≤ x ≤ b$ for all x in $a ≤ x ≤ b$, then the area between $y_1$ and $y_2$ from x = a to x = b is given by:

$$∫↙{a}↖{b} (y_1 - y_2)dx $$

The method of approximating the area under a curve by the sum of an infinite series of rectangles is known as Riemann sums, after the famous German mathematician, Georg Friedrich Bernhard Riemann, 1826 - 1866.

What is the area of the region bounded by the curves: $y = x^2$ and $y = 1 - x^2$?

First, find the points of intersection by equating the two equations:

$x^2=1-x^2$, ⇒ $x=±1/{√2}$

$A = ∫_{-1/{√2}}^{+1/{√2}} [f(x) - g(x)]dx = ∫_{-1/{√2}}^{+1/{√2}} [1-x^2 -x^2]dx = ∫_{-1/{√2}}^{+1/{√2}} [1-2x^2]dx = [x - 2/{3}x^3]_{-1/{√2}}^{+1/{√2}}$

$= 2/{3√2} - (-2/{3√2}) = 0.943$

The volume of a solid formed by rotating the area between a function and the x-axis through 360° is:

$$∫_a^bπy^2dx$$If a two-dimensional shape is rotated through 360°, a solid, 3D shape results. Integration allows us to calculate the volume of the solid formed by rotation.

A rectangle of height y and thickness dx rotated through 360° forms a disk, or narrow cylinder. This disk has a volume equal to the area of the circle it forms times the disk height (dx):

$$V_{disk} = πr^2h = πy^2 dx$$The volume of a 3D shape formed by a series of disks, from a to b on a curve, is:

$$∫↙{a}↖{b} πy^2dx$$Where $f(x) ≥ g(x)$, for all $x$ in the interval $[a, b]$, the volume of rotation formed by rotating the area between the two curves $2π$ radians about the $x$-axis in the interval $[a, b]$ is given by:

$$V = π∫_a^b (f(x))^2 - π∫_a^b (g(x))^2dx$$ $$= π∫_a^b [f(x)^2 - g(x)^2]dx$$Content © Renewable-Media.com. All rights reserved. Created : February 5, 2015

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Sir Paul Nurse is a British biologist and geneticist, who won a Nobel Prize for the discovery of controlling proteins in cell cycles. He is the current President of the Royal Society and Chief Executive and Director of the Francis Crick Institute.

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