To solve the problem, we need to evaluate the given integral condition: \(\int_0^x t f(t) \, dt = x^2 f(x)\) for \(x > 0\) and find \( f(6) \). First, differentiate both sides of the equation with respect to \(x\). The left side is an integral of the form \(\int_0^x u f(u) \, du\), which can be differentiated using the Leibniz rule for differentiation under the integral sign:
\(\frac{d}{dx}\left(\int_0^x tf(t) \, dt\right) = x f(x) + \int_0^x t \frac{d}{dx}f(t) \, dt\).
The right side of the equation differentiated with respect to \(x\) gives:
\(\frac{d}{dx}(x^2 f(x)) = 2x f(x) + x^2 f'(x)\).
Equating the derivatives of both sides:
\(x f(x) + \int_0^x t f'(t) \, dt = 2x f(x) + x^2 f'(x).\)
By comparing both sides, \(\int_0^x t f'(t) \, dt\) needs to satisfy \(x f(x) = 2x f(x) + x^2 f'(x)\).
This simplifies to:
\(0 = x f(x) + x^2 f'(x)\).
Rearranging terms gives us:
\(x f(x)(1 + x f'(x)/f(x)) = 0\).
Since \(x > 0\), it implies:
\(f(x) + x f'(x) = 0\), such that \(f(x)\) satisfies the differential equation \(f'(x)/f(x) = -1/x\).
This differential equation can be solved by separating variables:
\(\frac{d}{dx}\ln|f(x)| = -\frac{1}{x}\).
Integrate both sides:
\(\ln|f(x)| = -\ln|x| + C\) where \(C\) is the constant of integration.
\(|f(x)| = \frac{e^C}{x}\) gives our function \(f(x) = \frac{k}{x}\) where \(k = e^C\) is a constant.
Since \( f(2) = 3 \), substitute into the function:
\(f(2) = \frac{k}{2} = 3\) which implies \(k = 6\).
Therefore, the function is \(f(x) = \frac{6}{x}\).
Finally, we compute \( f(6) \): \(f(6) = \frac{6}{6} = 1\).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,