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
Yes 9 times pi is an irrational number because it can't be expressed as a fraction
3.1415925
Circumference: 2*pi*3 = 6*pi inches Area: pi*3 squared = 9*pi square inches
The integral of root(sin(x)) is -2 time the elliptic integral of the second order of .25(pi-2x) at 2. For this and other integrals, go to http://integrals.wolfram.com/index.jsp?expr=sqrt(sin(x))&random=false For more information on the elliptic integral functions, go to http://en.wikipedia.org/wiki/Elliptic_integral Hope this helps!
Same as any other function - but in the case of a definite integral, you can take advantage of the periodicity. For example, assuming that a certain function has a period of pi, and the value of the definite integral from zero to pi is 2, then the integral from zero to 2 x pi is 4.
The method to use is 'integration by parts'; set u =x; du=dx; dv = sin(pi x)dx; v = cos(pi x)/pi. so integral(u dv) = u*v - integral(v du) then repeat the process.
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
For example, by calculating the surface of a circle, using an integral.
Integrate the function for the curve, as normal, but the change the sign of the result. Be very careful that the curve is always on the same side of the x-axis between the limits of integration. If necessary, partition the integral. For example, to find the area between the x-axis and sin(x) between x=0 and x=3*pi, you will need Integral of sin(x) between 0 and pi, -[integral of sin(x) between pi and 2*pi] - this is where the curve is below the x-axis. +integral of sin(x) between 2*pi and 3*pi.
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.
22/7 pi = 2 * integral from 0 to infinity of (1 / (t2 + 1)) dt
Let y = sin 2x Then dy/dx = 2*cos 2x Also, when x = 0, y = 0 and when x = pi/6, y = sin(2*pi/6) = sin(pi/3) = sqrt(3)/2 Therefore, the integral becomes definite integral from 0 to √3/2 of 1/2*y3dy = difference between 1/2*(y4)/4 = (y4)/8 evaluated at √3/2 and 0. = (√3/2)4 /8 = 9/128 = 0.0703 approx.
1/3 [definite integral, interval [3,6]]((pi)r2) 1/3 (((pi)r3)/3) interval [3, 6] 1/3 [ (72(pi) - 9(pi) ] 1/3 [ 63(pi) ] 21pi Sorry about the set-up, could not figure out how to do all of the equations.
(3pi-7)/9 To verify this go to the link and enter integrate x cos(x)^3 from 0 to pi/2.
9*pi or 2*pi*4.5
9 is the 100th of pi