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Suppose $\forall x \in \mathbb{R}$, $I(x)=\displaystyle\int_{0}^{x}{\dfrac{(x-t)^2\exp(t)}{2}\,\mathrm dt}$.

Without calculating $I(x)$ how can I prove that:

  • $\forall x \ge 0$, $\quad0\le I(x) \le \dfrac{\exp(x)x^3}{6}$;
  • $\forall x\le0$, $\quad |I(x)|\le \dfrac{|x|^3}{6}$.
Lord_Farin
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pourjour
  • 1,070

2 Answers2

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HINT: $\exp(t)$ is an increasing function for $0<t<x$ for any $x \geq 0$, thus $$ |I(x)| = \left| \int_0^x \frac{(x-t)^2}{2} \exp(t) \mathrm{d}t \right| \leqslant \left| \int_0^x \frac{(x-t)^2}{2} \exp(x) \mathrm{d}t \right| = \exp(x) \left| \int_0^x \frac{(x-t)^2}{2} \mathrm{d}t \right| $$ You should be able to finish it off now

Sasha
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3

Just note that

$$ I(x)=\int_{0}^{x}{\dfrac{(x-t)^2\exp(t)}{2}dt}=\int_{0}^{x}{\dfrac{(t)^2\exp(x-t)}{2}dt}\longrightarrow (1) $$

$$ \implies |I(x)|\leq \frac{e^x}{2}\int_{0}^{x}t^2 dt =\dots,$$

since $e^{-t}\leq 1 $ on the interval $[0,x]$.

Added: Eq. $(1)$ follows from the property of the convolution

$$ \int_{0}^{x}f(x-t)g(t)dt = \int_{0}^{x}f(t)g(x-t)dt. $$

Mhenni Benghorbal
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