Q. 7 - Calculus

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.

Step-by-Step Solution

Answer

A(x) is concave down.

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

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