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The answer is ln (V) + C
What else can I help you with?
What is integral of dv?
∫dv = v + some_constant
What is the integral of x sin pi x?
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 of xlnxdx?
integration by parts. Let u=lnx, dv=xdx-->du=(1/x)dx, v=.5x^2. Integral of (xlnxdx)=lnx*.5x^2-integral of .5x^2(1/x)dx=lnx*.5x^2-integral of .5xdx=lnx*.5x^2-(1/6)x^3. That's it.
What is the integral of xsin2xdx?
∫ 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)
What is integral of dv?
∫dv = v + some_constant
What is the integral of x sin pi x?
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.
How do you convert divergent to surface integral?
To convert a divergence to a surface integral, you can use the Divergence Theorem, which states that for a vector field (\mathbf{F}) defined in a region (V) with a smooth boundary surface (S), the integral of the divergence of (\mathbf{F}) over (V) is equal to the flux of (\mathbf{F}) across (S). Mathematically, this is expressed as: [ \int_V (\nabla \cdot \mathbf{F}) , dV = \iint_S \mathbf{F} \cdot \mathbf{n} , dS ] where (\mathbf{n}) is the outward unit normal to the surface (S). Thus, you can transform a volume integral of divergence into a surface integral by applying this theorem.
Roman Numerals for 5-29-55?
V-IXXX-DV
How do you get acceleration?
a=dv/dt a=acceleration v=velocity t=time.
What do you do to get acceleration?
a=dv/dt a=acceleration v=velocity t=time.
How do you do acceleration?
a=dv/dt a=acceleration v=velocity t=time.
How do you calclate acceleration?
a=dv/dt a=acceleration v=velocity t=time.
How do you calulate acceleration?
a=dv/dt a=acceleration v=velocity t=time.
How do you caluclate acceleration?
a=dv/dt a=acceleration v=velocity t=time.
Integral of xlnxdx?
integration by parts. Let u=lnx, dv=xdx-->du=(1/x)dx, v=.5x^2. Integral of (xlnxdx)=lnx*.5x^2-integral of .5x^2(1/x)dx=lnx*.5x^2-integral of .5xdx=lnx*.5x^2-(1/6)x^3. That's it.
Integration of xtanx?
To integrate ( x \tan(x) ), we can use integration by parts. Let ( u = x ) and ( dv = \tan(x) , dx ). This gives ( du = dx ) and ( v = -\ln|\cos(x)| ). Applying the integration by parts formula ( \int u , dv = uv - \int v , du ), we obtain: [ \int x \tan(x) , dx = -x \ln|\cos(x)| + \int \ln|\cos(x)| , dx + C ] The integral ( \int \ln|\cos(x)| , dx ) does not have a simple closed form, so the final result may be expressed in terms of this integral along with the logarithmic term.
