Q. 30 - Calculus

Question

For each area accumulation function A in Exercises 27–30, 

(a) illustrate A(2) graphically, 

(b) calculate A(2) and A(5), and 

(c) find an explicit elementary formula for A(x).

$A\left(x\right)={\int }_{-x}^x{t}^3.dt$

Step-by-Step Solution

Answer

(a) The function A(2) is:

(b) The value of $A\left(2\right)=0 and \left(A\right(5)=0$.

(c) The explicit elementary formula is $A\left(x\right)=0$.

Part (a) Step 1. Given Information.

The function:

$A\left(x\right)={\int }_{-x}^x{t}^3.dt$

Part (a) Step 2. Graph the function.

Graph the function, $A\left(2\right)={\int }_{-2}^2{t}^3.dt$

So graph the function $A\left(2\right)={\int }_{-2}^2{t}^3.dt$ between the points $x=-2 to x=2$.

Part (b) Step 1. Calculate A(2).

$$\begin{gathered} A\left(x\right)={\int }_{-x}^x{t}^3.dt \\ A\left(2\right)={\int }_{-2}^2{t}^3.dt \\ ={\left[\frac{t^4}{4}\right]}_{-2}^2 =\frac{16}{4}-\frac{16}{4} \\ =0 \end{gathered}$$

Part (b) Step 2. Calculate A(5).

$$\begin{gathered} A\left(x\right)={\int }_{-x}^x{t}^3.dt \\ A\left(5\right)={\int }_{-5}^5{t}^3.dt \\ ={\left[\frac{t^4}{4}\right]}_{-5}^5 \\ =\frac{625}{4}-\frac{625}{4} \\ =0 \end{gathered}$$

Part (c) Step 1. Find an explicit formula.

Find an explicit formula for the given function. 

$$\begin{gathered} A\left(x\right)={\int }_{-x}^x{t}^3.dt \\ ={\left[\frac{t^4}{4}\right]}_{-x}^x \\ =\frac{x^4}{4}-\frac{x^4}{4} \\ =0 \end{gathered}$$