Where you refer to a particular integral I will assume you mean a definite integral. To illustrate why there is no constant of integration in the result of a definite integral let me take a simple example. Consider the definite integral of 1 from 0 to 1. The antiderivative of this function is x + C, where C is the so-called constant of integration. Now to evaluate the definite integral we calculate the difference between the value of the antiderivative at the upper limit of integration and the value of it at the lower limit of integration: (1 + C) - (0 + C) = 1 The C's cancel out. Furthermore, they will cancel out no matter what the either antiderivatives happen to be or what the limits of integration happen to be.
The answer depends on what group or field the function is defined on. In the complex plane, the range is the complex plane. If the domain is all real numbers and the radical is an odd root (cube root, fifth root etc), the range is the real numbers. Otherwise, it is the complex plane. If the domain is non-negative real numbers, the range is also the real numbers.
First, find the area under the curve y = 1/x2, with boundary lines x = 1 and x = 6, by calculating the integral of 1/x2 with lower limit 1, and upper limit 6. Then divide it by 2. (6)integral(1) of (1/x2) dx = (6)integral(1) of (x--2) dx = -x-1|(6),(1) = -1/x|(6)(1) = -1/6 +1 = 5/6. Thus, the half of the area under the curve is 5/12.
You could find an inclined plane in a skate park because of all the ramps.
38; 39; 40!
All complex numbers are part of the "complex plane", so none of them is farther than others.
find the area of bounded by the two curves. y=9-x
Where you refer to a particular integral I will assume you mean a definite integral. To illustrate why there is no constant of integration in the result of a definite integral let me take a simple example. Consider the definite integral of 1 from 0 to 1. The antiderivative of this function is x + C, where C is the so-called constant of integration. Now to evaluate the definite integral we calculate the difference between the value of the antiderivative at the upper limit of integration and the value of it at the lower limit of integration: (1 + C) - (0 + C) = 1 The C's cancel out. Furthermore, they will cancel out no matter what the either antiderivatives happen to be or what the limits of integration happen to be.
The answer depends on what group or field the function is defined on. In the complex plane, the range is the complex plane. If the domain is all real numbers and the radical is an odd root (cube root, fifth root etc), the range is the real numbers. Otherwise, it is the complex plane. If the domain is non-negative real numbers, the range is also the real numbers.
8
First, find the area under the curve y = 1/x2, with boundary lines x = 1 and x = 6, by calculating the integral of 1/x2 with lower limit 1, and upper limit 6. Then divide it by 2. (6)integral(1) of (1/x2) dx = (6)integral(1) of (x--2) dx = -x-1|(6),(1) = -1/x|(6)(1) = -1/6 +1 = 5/6. Thus, the half of the area under the curve is 5/12.
integral.
111
Use a line integral.
One can find many examples of integral tables online. Sites such as Mathwords, Math2org, Cobalt and SOSMath have many examples available for use as well as instructions on how to use them.
The integral function of calculus is the method for determining the area under a curve. The limiting chord process is the "simple" math understanding required to learn the "complex" function of "integration". BTW: the derivative function is a "cousin" of the integral function which is used to determine the slope of curve at a given point.
To Find The Plane,You Need To Go Left From The Museum, Past The Fly By Night Building And The Plane Will Be There.