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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
