Q. 27 - 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).

$A\left(x\right)={\int }_1^x(t^2+1)dt$

Step-by-Step Solution

Answer

(a) The function A(2) is:

(b) The value of $A\left(2\right)=\frac{10}{3} and A\left(5\right)=\frac{136}{3}$.

(c) The explicit elementary formula is $A\left(x\right)=\frac{x^3}{3}+x-\frac{4}{3}$.

Part (a) Step 1. Given Information.

The function:

$A\left(x\right)={\int }_1^x(t^2+1)dt$

Step 2. Graph the function.

Graph the function,

$A\left(2\right)={\int }_1^2(t^2+1).dt$

So graph the function $t^2+1$ between the points $x=1 to x=2$.

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

$$\begin{gathered} A\left(2\right)={\int }_1^2(t^2+1).dt \\ ={[\frac{t^3}{3}+t]}_1^2 \\ =[\frac{2^3}{3}+2]-[\frac{1^3}{3}+1] \\ =\frac{8}{3}+2-\frac{1}{3}-1 \\ =\frac{7}{3}+1 \\ =\frac{10}{3} \end{gathered}$$

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

$$\begin{gathered} A\left(25\right)={\int }_1^5(t^2+1).dt \\ ={[\frac{t^3}{3}+t]}_1^5 \\ =[\frac{5^3}{3}+5]-[\frac{1^3}{3}+1] \\ =\frac{125}{3}+5-\frac{1}{3}-1 \\ =\frac{124}{3}+4 \\ =\frac{136}{3} \end{gathered}$$

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

Find an explicit formula for the given function.

$$\begin{gathered} A\left(x\right)={\int }_1^x(t^2+1)dt \\ ={[\frac{t^3}{3}+t]}_1^x \\ =[\frac{x^3}{3}+x]-[\frac{1}{3}+1] \\ =\frac{x^3}{3}+x-\frac{4}{3} \end{gathered}$$