Question

Consider the function

$A\left(x\right)={\int }_a^{x}f\left(t\right)dt.$

Where f is positive and decreasing for x>0

Find A(x) and also to argue that if f(x) is decreasing then A(x) must be

concave down.

Short answer

A(x) is concave down.

Step-by-step solution

Step 1. Given information

$A\left(x\right)={\int }_a^{x}f\left(t\right)dt.$Consider the function

Where f is positive and decreasing for x>0

Step 2. Calculation

$$\begin{gathered} A\left(x\right)=\frac{d}{dx}\left(\frac{d}{dx}\right({\int }_a^{x}f\left(t\right)dt\left)\right) \\ =\frac{d}{dx}\left(f\right(x\left)\right) \\ ={\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right) \\ \mathrm{As}\mathrm{f}\mathrm{is}\mathrm{decreasing}\mathrm{and}\mathrm{so}\mathrm{thus}\mathrm{f}\text{'}\mathrm{is}\mathrm{negative}\mathrm{which}\mathrm{implies}\mathrm{that}{\mathrm{A}}^{*}\left(\mathrm{x}\right)\mathrm{is}\mathrm{also}\mathrm{negative}. \\ \mathrm{Hence},\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}{\mathrm{A}}^{*}\left(\mathrm{x}\right)\mathrm{will}\mathrm{be}\mathrm{negative}. \\ \mathrm{Agraph}\mathrm{with}\mathrm{negative}\mathrm{rate}\mathrm{is}\mathrm{always}\mathrm{concave}\mathrm{downward}.\mathrm{Therefore},\mathrm{A}\left(\mathrm{x}\right)\mathrm{is}\mathrm{concave}\mathrm{down}. \end{gathered}$$