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Integrate(0->t) (2-2x) dx is the integral correct (Integrate(0->t) (2-2x) dy would be different, you must state integrate with respect to what otherwise it can be anything)
So integrals preserves sum and product with constants. i.e. Integrate (2-2x) dx = Integrate 2 dx - 2integrate x dx = 2x - x^2 + C
By Fundamental Theorem of Calculus, take any anti-derivative, say C = 0 would be fine, and Integral(0->t)(2-2x) dx = (2x-x^2)|(0->t) = (2t-t^2) - 0 = 2t-t^2
It is a special case of the Second Fundamental Theorem of Calculus -- integral(0->t) f(x) dx is an anti-derivative of f(x).
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Proof of Clausius inequality for reversible process?
Clausius Inequality is not only for reversible process Since overall (system + surrounding) entropy always increase. (∆s) system + surrounding ≥ 0 For a cyclic process (∆s) system = 0 So (∆s) surrounding ≥ 0 Note that in cyclic integral of (dQ/T) dQ is positive for heat entering system form surrounding. So for surrounding change in entropy after a cyclic process is given by cyclic integral of (-dQ/T) which is ≥ 0 Thus cyclic integral of (dQ/T) ≤ 0
What is the integral of 2 1-x dx with limits 0 to t?
I think you meant the integral of 21-x dx, or as I write it ,/`21-x dx.Use u-substitution where u = 1 - x. Note that the derivative of this equation is du = -1 dx, so we know dx = - du.Rewriting the original integral in terms of u, integrating, then putting it back in terms of x gives us:,/` 21-x dx,/` 2u (-du)- ,/` 2u du- 2u / ln(2)- 21-x / ln(2)Note that the second to last line can be found in any integral table, and ln denotes natural log.Now the bounds go from x = 0 to x = t, so:[- 21-t / ln(2)] - [- 21-0 / ln(2)]- 21-t / ln(2) + 2 / ln(2)(-21-t + 2)/ln(2)
An airplane must reach a speed of 195 miles hour to take off if the the runway is 456 meter long what is the minimum value of the acceleration that will allow the airplane to take off successfully?
The change in velocity is the integral of acceleration with respect to time. Assuming a constant acceleration, thenv = integral [a dt] = a t + v0The change in distance is the integral of velocity with respect to time:s = integral (v dt) = integral [(at + v0) dt] = 1 /2 a t^2 + v0 t + s0Since the airplane is taking off from standing still at the start of the runway, s0 = 0 and v0 = 0.s = 1/2 a t^2We know that the at the end of the runway (456 m), the velocity must be 195 mph or 87.2 m/s, and this is v = at for constant acceleration. Plugging this in, we get:456 m = 1/2 (87.2 m/s) tSolving for t we get t = 10.5 secondsPlugging this back into the equation for s, then456 m = 1/2 a (10.5 s)^2Solving for a:a = 8.27 m/s^2
How do you write t less than 0?
t < 0
H equals 16t2 plus 64t?
16t(t + 4) is the factorization Usually it's set = to 0 16t (t+4) =0 So, either 16t = 0 or t + 4 = 0 t = 0 or t = -4
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
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]
Derive the equation for average value of DC?
The equation for the average over time T is integral 0 to T of I.dt
Proof of Clausius inequality for reversible process?
Clausius Inequality is not only for reversible process Since overall (system + surrounding) entropy always increase. (∆s) system + surrounding ≥ 0 For a cyclic process (∆s) system = 0 So (∆s) surrounding ≥ 0 Note that in cyclic integral of (dQ/T) dQ is positive for heat entering system form surrounding. So for surrounding change in entropy after a cyclic process is given by cyclic integral of (-dQ/T) which is ≥ 0 Thus cyclic integral of (dQ/T) ≤ 0
What is the integral of 2 1-x dx with limits 0 to t?
I think you meant the integral of 21-x dx, or as I write it ,/`21-x dx.Use u-substitution where u = 1 - x. Note that the derivative of this equation is du = -1 dx, so we know dx = - du.Rewriting the original integral in terms of u, integrating, then putting it back in terms of x gives us:,/` 21-x dx,/` 2u (-du)- ,/` 2u du- 2u / ln(2)- 21-x / ln(2)Note that the second to last line can be found in any integral table, and ln denotes natural log.Now the bounds go from x = 0 to x = t, so:[- 21-t / ln(2)] - [- 21-0 / ln(2)]- 21-t / ln(2) + 2 / ln(2)(-21-t + 2)/ln(2)
What is the integral of sin t dt with the interval x 0?
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.
In line integral and surface integral you need a vector function f but integrand were actually scalar function why?
You don't need a vector function. A scalar function over a line or surface, and often this is easier than a line or surface integral of a vector function.When you do a line or surface integral of a vector function, you're really integrating the dot product of the vector function with the tangent to the line (for a line integral), or integrating the dot product of the vector function with the normal to a surface (for a surface integral). In both cases, the result of a dot product is just a scalar that you end up integrating over the line or surface. When you do a line or surface or integral of a scalar function you're halfway there already because you're skipping the vector dot product!Line integrals:If you think about it, the usual integration a function f(x) from a to b is equivalent to the line integral of the scalar function f(x) in the xy plane (f simply doesn't depend on y) along the straight line segment from (a, 0) to (b, 0). A more general line integral just extends this idea to an integral over a non-straight line.To do a general line integral of a scalar function f(x) (where x is a vector x = ) along a curve, you must parameterize the curve. Typically the parameterization is denoted by a vector-valued function, r(t) = , of a single parameter, t. Typically bounds would be given for t, such as a
What is the laplace transform of log function?
The Laplace transform is a widely used integral transform in mathematics with many applications in physics and engineering. It is a linear operator of a function f(t) with a real argument t (t ≥ 0) that transforms f(t) to a function F(s) with complex argument s, given by the integral F(s) = \int_0^\infty f(t) e^{-st}\,dt.
An airplane must reach a speed of 195 miles hour to take off if the the runway is 456 meter long what is the minimum value of the acceleration that will allow the airplane to take off successfully?
The change in velocity is the integral of acceleration with respect to time. Assuming a constant acceleration, thenv = integral [a dt] = a t + v0The change in distance is the integral of velocity with respect to time:s = integral (v dt) = integral [(at + v0) dt] = 1 /2 a t^2 + v0 t + s0Since the airplane is taking off from standing still at the start of the runway, s0 = 0 and v0 = 0.s = 1/2 a t^2We know that the at the end of the runway (456 m), the velocity must be 195 mph or 87.2 m/s, and this is v = at for constant acceleration. Plugging this in, we get:456 m = 1/2 (87.2 m/s) tSolving for t we get t = 10.5 secondsPlugging this back into the equation for s, then456 m = 1/2 a (10.5 s)^2Solving for a:a = 8.27 m/s^2
How do you draw a momentum -time graph for force-time graph?
The momentum-time graph is the integral of the force-time graph. that is, it is the area under the curve of the f-t graph.The momentum-time graph is the integral of the force-time graph. that is, it is the area under the curve of the f-t graph.The momentum-time graph is the integral of the force-time graph. that is, it is the area under the curve of the f-t graph.The momentum-time graph is the integral of the force-time graph. that is, it is the area under the curve of the f-t graph.
What is the ans of integration of x with respect to t?
The indefinite integral of x dt is xt
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).
