Find the curves y=f(x) such that the x -coordinate of the…

Find the curves \(y=f(x)\) such that the \(x\)-coordinate of the centroid of the curvilinear trapezoid bounded by the curve, the coordinate axes, and the line \(x=a\) equals \(\dfrac{3a}{4}\) for every \(a>0\).

1. The governing integral identity

The centroid condition \(\bar x = \dfrac{\int_0^a xy\,dx}{\int_0^a y\,dx}=\dfrac{3a}{4}\) for every \(a\) rearranges to \(4\!\int_0^a xy\,dx = 3a\!\int_0^a y\,dx\quad(1)\).

2. First differentiation (Leibniz rule)

Sympy's \(\texttt{diff}\) applied to \(\texttt{Integral}(t\,f(t),(t,0,a))\) and to \(\texttt{Integral}(f(t),(t,0,a))\) with respect to \(a\) returns exactly \(a\,f(a)\) and \(f(a)\) respectively — confirming the Leibniz rule for a variable upper limit as used by the source. Differentiating (1) then gives \(4af(a) = 3\!\int_0^a f\,dx+3af(a)\), i.e. \[ \int_0^a y\,dx = \frac{ay}{3}.\qquad(2) \]

3. Second differentiation

Differentiating (2) the same way gives \(y = \tfrac{y}{3}+\tfrac{a}{3}\dfrac{dy}{da}\), i.e. \(2y = a\,dy/da\), a separable equation \(\dfrac{dy}{y}=\dfrac{2\,da}{a}\) whose solution is \(y=ka^2\), i.e. (renaming the integration variable back to \(x\)) \(y=Cx^2\).

4. Verify the intermediate relations and the final condition directly

Substituting \(y=kx^2\) into Sympy's \(\texttt{integrate}\), relation (2) \(\int_0^a y\,dx = \tfrac{a\,y(a)}{3}\) checks out exactly (\(\texttt{simplify}\) of the difference is \(0\)), as does the original relation (1). Finally, computing \(\int_0^a y\,dx = \tfrac{ka^3}{3}\) and \(\int_0^a xy\,dx=\tfrac{ka^4}{4}\) directly with Sympy and forming their ratio gives \(\bar x = \tfrac{3a}{4}\) exactly — reproducing and confirming the source's own closing verification.