Q. 10 - Calculus

Question

Consider the function

$A\left(x\right) = {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }} \cos x dt; B\left(x\right) ={\int }_{\mathrm{\pi }}^x \cos x dt$

Explain why their derivatives differ by a constant in three  different ways .

Step-by-Step Solution

Answer

The derivative of  both function differs by constant $A\left(x\right) \&B\left(x\right)$

The derivative of  both function differs by constant showing algebrically  $A\left(x\right) -B\left(x\right)$

The third way is depicting graphically.

Step 1. Given

The given function is $A\left(x\right) = {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }} \cos x dt; B\left(x\right) ={\int }_{\mathrm{\pi }}^x \cos x dt$$A\left(x\right) = {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }} \cos x dt; B\left(x\right) ={\int }_{\mathrm{\pi }}^x \cos x dt$

Step 2. Explaining the first way

The first way is by comparing the derivatives of $A\left(x\right) - B\left(x\right)$

$$\begin{gathered} \mathrm{The} \mathrm{derivative} \mathrm{of} \mathrm{A}\left(\mathrm{x}\right) \mathrm{is} \mathrm{A}\text{'}\left(\mathrm{x}\right) =\mathrm{cosx} \\ \mathrm{The} \mathrm{derivative} \mathrm{of} B\left(\mathrm{x}\right) \mathrm{is} \mathit{B}\text{'}\left(\mathrm{x}\right) =\mathrm{cosx} \end{gathered}$$

The both derivative differs by constant.

Step 3, Explaining the second way (algebriacally)

$$\begin{gathered} A\left(x\right)-B\left(x\right) \\ ={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\cos x dt-{\int }_x^{\mathrm{\pi }}\cos x dt \\ ={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\cos x dt \\ \mathrm{It} \mathrm{is} \mathrm{known} \mathrm{that} {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\cos x dt is cons\tan t \\ \mathrm{Hence} \mathrm{derivative} \mathrm{differ} \mathrm{by} \mathrm{a} \mathrm{constant} \mathrm{showing} \mathrm{algebraically} \mathrm{that} A\left(x\right)-B\left(x\right) is cons\tan t \end{gathered}$$

Step 4. Third way showing graphically

$$\begin{gathered} A\left(x\right)-B\left(x\right)={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\cos xdt \\ \mathrm{Which} \mathrm{means} \mathrm{that} \mathrm{it} \mathrm{is} \mathrm{the} \mathrm{signed} \mathrm{area} \mathrm{under} \mathrm{the} \mathrm{graph} \mathrm{of} y=\cos x \\ from x=-\mathrm{\pi } \mathrm{to} \mathrm{x}=\mathrm{\pi } \end{gathered}$$