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Well, sweetheart, the integral of dV/V is simply ln|V| + C, where C is the constant of integration. So, in other words, the integral of dV/V is the natural logarithm of the absolute value of V, plus some boring constant. Math can be a real snoozefest, but hey, at least now you know the answer!
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BeauThe integral of dV/V is the natural logarithm of the absolute value of V, denoted as ln|V|, plus a constant of integration, often denoted as +C. This integral arises in calculus when integrating functions involving inverse relationships, such as exponential growth or decay. The result represents the accumulation of small changes in V over a given interval, providing a mathematical tool to analyze and model various phenomena in science and engineering.
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What is integral of dv?
∫dv = v + some_constant
What is the integral of x sin pi x?
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Integral of xlnxdx?
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∫ xsin(2x) dx = (-1/2)xcos2x + (1/4)sin2x You get this by using Integration by Parts. An integral in the form ∫udv can be written as uv-∫vdu In the case of your problem u=x, du=1, dv=sin2x, v=(-1/2)cos2x <--You get v by integrating dv Using the formula ∫udv = uv- ∫vdu and by plugging in what has been defined above you get ∫xsin(2x)dx = (-1/2)xcos2x - ∫(-1/2)cos2x(1) By integrating ∫(-1/2)cos2x, you get (-1/4)sin2x. When you plug that back in, you get ∫xsin2xdx=(-1/2)xcos2x-(-1/4)sin2x or just simply ∫xsin(2x)dx = (-1/2)xcos(2x) + (1/4)sin(2x)
What is the Integral of 23x e2x dx?
This browser is pathetic for mathematical answers but here's the best that I can do:Let u = 23x therefore du/dx = 23let dv/dx = e^(2x) therefore v = 1/2*e^(2x)then, integrating by parts,I = I(u*dv/dx) dx = u*v - I(du/dx*v) dx= 23x*(1/2)*e^(2x) - I(23*(1/2)*e^(2x) dx= 23/2*xe^(2x) - 23/2*I(e^(2x)) dx= 23/2*xe^(2x) - 23/2*(1/2)*e^(2x)= 23/2*xe^(2x) - 23/4*e^(2x)or 23/4*e^(2x)*(2x - 1)
