Question

Let $f:[0,27]\to [\frac{1}{3},6]$ be a differentiable function such that $f^{\prime }\left(x\right)<0\forall x\in D_f$. If $\underset{0}{\overset{27}{\int }}x{f}^{\prime }\left(x\right)dx=\lambda -3\underset{0}{\overset{3}{\int }}{x}^{2}f\left(x^3\right)dx,$ then the minimum value of $\lambda$ is
(Note: $D_f$ denotes the domain of the function)

Answer 1

$\underset{0}{\overset{27}{\int }}x{f}^{\prime }\left(x\right)dx=\lambda -3\underset{0}{\overset{3}{\int }}{x}^{2}f\left(x^3\right)dx\Rightarrow \lambda =\underset{0}{\overset{27}{\int }}x{f}^{\prime }\left(x\right)dx+3\underset{0}{\overset{3}{\int }}{x}^{2}f\left(x^3\right)dx$

Putting $x^3=t\Rightarrow 3x^{2}dx=dt$
$\Rightarrow \lambda =\underset{0}{\overset{27}{\int }}x{f}^{\prime }\left(x\right)dx+\underset{0}{\overset{27}{\int }}f\left(t\right)dt\Rightarrow \lambda =\underset{0}{\overset{27}{\int }}x{f}^{\prime }\left(x\right)dx+\underset{0}{\overset{27}{\int }}f\left(x\right)dx\Rightarrow \lambda =\underset{0}{\overset{27}{\int }}\left[f\right(x)+x{f}^{\prime }(x\left)\right]dx\Rightarrow \lambda ={\left[xf\right(x\left)\right]}_0^{27}\Rightarrow \lambda =27f\left(27\right)$

As $f^{\prime }\left(x\right)<0\forall x\in D_f$, so $f\left(x\right)$ is monotonically decreasing function, then
$\left[f\right(27){]}_{min}=\frac{1}{3}$
Therefore, ${\lambda }_{min}=27\times \frac{1}{3}=9$