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Let g(x) = interval [0, x] of sin t dt, and f(t) = sin t.
Since f(t) is a continuous function, the part one of the Fundamental Theorem of Calculus gives, g'(x) = sin x = f(x) (the original function).
If you are interested in the interval [x, 0] of sin t dt, then just put a minus sign in front of the integral and interchange places of 0 and x. So that,
g(x) = interval [x, 0] of sin t dt = -{ interval [0, x] of sin t dt}, then g'(x) = - sin x.
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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.
What is the value of y sin θ when θ 0?
sin(0)=0, therefore ysin(0)=0
What is the integral of 0?
The integral of 0 is some constant C. You can solve for this constant by using boundry conditions if there are any given; otherwise, just put C.
What is limit as x approaches 0 of sin squared x by x?
lim(x->0) of sin(x)^2/x we use L'Hospital's Rule and derive the top and the bottomd/dx(sin(x)^2/x)=2*sin(x)*cos(x)/1lim(x->0) of 2*sin(x)*cos(x)=2*0*1=0
If the derivative of a function equals xsquared - 2divided byx on which intervals is f decreasing?
f(x) is decreasing on the interval on which f'(x) is negative. So we want: (x2-2)/x<0 For this to be true either the numerator or the denominator (but not both) must be negative. On the interval x>0, the numerator is negative for 0<x<sqrt(2) and the denominator is positive for all x>0. On the interval x<0, the denominator is negative for all values on this interval. The numerator is positive on this interval for x<-sqrt(2). So, f' is negative (and f is decreasing) on the intervals: (-infinity, -sqrt(2)), (0, sqrt(2))
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.
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]
Sin xcos x sin x - cos x - 1 0 Then on the interval 0 2p x?
umm who knows
What math function gives you pi?
22/7 pi = 2 * integral from 0 to infinity of (1 / (t2 + 1)) dt
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 the definite integral from 0 to pi divided by 6 of sin cubed 2x cos 2x dx?
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.
How determine the area of curve under the x-axis?
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.
Can you have a zero velocity and zero displacement?
Yes, dD/dt = d0/dt = 0 thusDisplacement D=0 and Velocity dD/dt=d0/dt = o.
What are allowable probabilities?
A probability must be a real number in the interval [0, 1]. The sum (or integral) of the probabilities over all possible values must be 1.
How do you solve sin 2x equals sin x over 2?
they do have calculators for these questions you knowsin 2x = (sin x)/22 sin x cos x - (1/2)sin x = 02 sin x(cos x - 1/4) = 02 sin x = 0 or cos x - 1/4 = 0sin x = 0 or cos x = 1/4in the interval [0, 360)sin x = 0, when x = 0, 180cos x = 1/4, when x = 75.52, 284.48Check:
Tentukan panjang busur dari kurva (y+1)^2=4x^2 antara (0,-1) dan (-2,3)?
Untuk menentukan panjang busur dari kurva antara dua titik yang diberikan, kita dapat menggunakan rumus integral. Dalam kasus ini, kita perlu mengubah persamaan kurva menjadi parameterisasi melalui parameter t. Langkah-langkahnya adalah sebagai berikut: Memulai dengan persamaan kurva: (y + 1)^2 = 4x^2 Ubah persamaan tersebut menjadi parameterisasi menggunakan parameter t: x = t y = 2t^2 - 1 Hitung turunan pertama dari parameterisasi tersebut: dx/dt = 1 dy/dt = 4t Gunakan rumus integral untuk menghitung panjang busur: S = ∫√(dx/dt)^2 + (dy/dt)^2 dt Hitung integral tersebut antara batas-batas yang diberikan, yaitu dari t = 0 hingga t = -2: S = ∫[0,-2] √(1^2 + (4t)^2) dt Terapkan rumus integral: S = ∫[0,-2] √(1 + 16t^2) dt Langkah selanjutnya adalah menyelesaikan integral tersebut. Namun, integral ini melibatkan akar kuadrat, sehingga harus diselesaikan menggunakan teknik integral yang lebih lanjut, seperti substitusi atau metode numerik. Karena itu, dalam kasus ini, lebih baik menggunakan metode numerik atau menggunakan perangkat lunak matematika seperti Mathematica atau MATLAB untuk menyelesaikan integral dan menghitung panjang busur secara akurat antara kedua titik yang diberikan, yaitu (0, -1) dan (-2, 3).
What are the two basic properties of all probability distributions?
The probability of any event lies in the interval [0, 1]. The sum (or integral) over all possible outcomes is 1.
