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What is the relation between definite integrals and areas?
Consider the integral of sin x over the interval from 0 to 2pi. In this interval the value of sin x rises from 0 to 1 then falls through 0 to -1 and then rises again to 0. In other words the part of the sin x function between 0 and pi is 'above' the axis and the part between pi and 2pi is 'below' the axis. The value of this integral is zero because although the areas enclosed by the parts of the function between 0 and pi and pi and 2pi are the same the integral of the latter part is negative. The point I am trying to make is that a definite integral gives the area between a function and the horizontal axis but areas below the axis are negative. The integral of sin x over the interval from 0 to pi is 2. The integral of six x over the interval from pi to 2pi is -2.
How do you use an arc length integral to show the length of the circle of radius r?
The integral from 0 to 2 pi of your constant value r will equal the circumference. This will be equal to 2*pi*r. This can be derived because of the following: Arc length = integral from a to b of sqrt(r^2-(dr/dtheta)^2) dtheta. By substituting the equation r = a constant c, dr/dtheta will equal 0, a will equal 0, and b will equal 2pi (the radians in a circle). By substitution, this becomes the integral from 0 to 2 pi of sqrt(c^2 + 0)dtheta, which leads us back to the original formula.
What is the indefinite integral of 3sinx-5cosx?
8
What is the convulusion integral?
Do you mean the Convolution Integral?
What is the integral of 2 times 1-x dx with limits 0 to t?
,/` 2(1 - x) dx,/` 2 - 2x dx2x - x2 ...evaluated from 0 to t gives us...2t - t2 - [2(0) - (0)2]2t - t2
What is the ISBN of The Integral Trees?
The ISBN of The Integral Trees is 0-345-31270-8.
What is the integral from 0 to pi over 6 sine 2x dx?
Integral from 0 to pi 6sin2xdx: integral of 6sin2xdx (-3)cos2x+c. (-3)cos(2 x pi) - (-3)cos(2 x 0) -3 - -3 0
What are the different examples of integral exponents?
... -3, -2, -1, 0, 1, 2, 3, ...In summary, any integer that you use as an exponent is an "integral exponent".... -3, -2, -1, 0, 1, 2, 3, ...In summary, any integer that you use as an exponent is an "integral exponent".... -3, -2, -1, 0, 1, 2, 3, ...In summary, any integer that you use as an exponent is an "integral exponent".... -3, -2, -1, 0, 1, 2, 3, ...In summary, any integer that you use as an exponent is an "integral exponent".
What is the integral of 1 divided by x squared?
The indefinite integral of (1/x^2)*dx is -1/x+C.
Integration of y from 0 to infinity?
Gaussian integral method is used in Integration of y from 0 to infinity.
What is the relation between definite integrals and areas?
Consider the integral of sin x over the interval from 0 to 2pi. In this interval the value of sin x rises from 0 to 1 then falls through 0 to -1 and then rises again to 0. In other words the part of the sin x function between 0 and pi is 'above' the axis and the part between pi and 2pi is 'below' the axis. The value of this integral is zero because although the areas enclosed by the parts of the function between 0 and pi and pi and 2pi are the same the integral of the latter part is negative. The point I am trying to make is that a definite integral gives the area between a function and the horizontal axis but areas below the axis are negative. The integral of sin x over the interval from 0 to pi is 2. The integral of six x over the interval from pi to 2pi is -2.
What is the integretion of modxdx?
mod x, or |x| is actually a conjunction of two functions: 1) x = -x, for x < 0 2) x = x, for x >= 0. Whenever you're calculating integral of |x|, you have to consider those two functions, for example: integral of |x| from -5 to 4 by dx is a sum of integrals of -x from -5 to 0 by dx and integral of x from 0 to 4 by dx.
RMS of non sinusoidal?
Vrms=sqrt[1/T * integral(v^2(t)dt, 0,t] Irms=sqrt[1/T * integral(i^2(t)dt, 0,t]
What is the area of the region between y equals x2 -5x and the x axis a from 0 to5 b from 0 to 6 cfrom-2 to5?
y = x2 - 5xThe integral of [ y dx ] = x3/3 - 5x2/2 + CAt x=0, the integral = 0At x=5, the integral = 125/3 - 125/2 = -20.8333At x=6, the integral = 72 - 90 = -18At x = -2, the integral = -8/3 - 10 = -12.6666a). from 0 to 5, the area is -20.8333 (20.8333 below the x-axis)b). from 0 to 6, the area is -18.c). from -2 to +5, the area is (-20.8333 + 12.6666) = +9.8333
What is an integral that equals 2012 and for all the smart people out there can you give it as a formula?
For the integral to equal 2012, we need the derivative of 2012. Because this is zero, we could assume that there are no values that would give you 2012. However, if you integrate 0, you get a constant, and therefore, because we can choose this constant to be whatever we require, the integral of 0 could possibly equal 2012. However, if you are (more likely) required to find the integral of 2012, its 2012x, or the derivative of 2012 is as mentioned earlier, 0
What is the integral of cosx divided by sinx plus cosx from 0 to 2pi?
The Integral diverges. It has singularities whenever sin(x)+cos(x)=0. Singularities do not necessarily imply that the integral goes to infinity, but that is the case here, since the indefinite integral is x/2 + 1/2 Log[-Cos[x] - Sin[x]]. Obviously this diverges when evaluated at zero and 2pi.
What is integral of 2x?
The integral of 2x is x^2+c, where c is a constant. If this is a definite integral, meaning that the limits of integration are known, then c=0. If this is an indefinite integral, meaning the limits of integration are unknown, then c should either be left as is or solved for using an initial condition.
