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91Ó°ÊÓ

Evaluate d2dx2∫0x∫0xfs,tdsdt .

Short Answer

Expert verified

Thus, .d2dx2∫0x∫0xfs,tdsdt=2fx+∫0x∂f∂xdt+∫0x∂f∂xds

Step by step solution

01

Determine the integral value.

d2dx2∫0x∫0xfs,tdsdt=ddxddx∫0xds∫0xfs,tdt=ddx∫0xfx,tdt+∫0xds∂∂x∫0xfs,tdt=ddx∫0xfx,tdt+∫0xfs,x+∫0x∂fs,t∂xdt

So, we get,

d2dx2∫0x∫0xfs,tdsdt=ddx∫0xfx,tdt+∫0xfs,xds=ddx∫0xfx,tdt+ddx∫0xfs,xds

02

Determine the Solution using formulas.

Now we use the formulas are,

ddx∫0xfx,tdt=fx+∫0x∂f∂xdtddx∫0xfs,xds=fx+∫0x∂f∂xds

Substituting these formulas,

d2dx2∫0x∫0xfs,tdsdt=fx+∫0x∂f∂xdt+fx+∫0x∂f∂xds=2fx+∫0x∂f∂xdt+∫0x∂f∂xds

So, the solution is,

d2dx2∫0x∫0xfs,tdsdt=2fx+∫0x∂f∂xdt+∫0x∂f∂xds

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