Q.32 - Calculus

Question

For each function f and interval [a, b] in Exercises 27–33, use the given approximation method to approximate the signed area between the graph of f and the x-axis on [a, b]. Determine whether each of your approximations is likely to be an over-approximation or an under-approximation of the actual area.  

$f\left(x\right)=9-x^2, [a,b]=\left[0, 5\right], n=5$ with

(a) midpoint sum                (b) lower sum 

Step-by-Step Solution

Answer

Using approximation method to approximate the signed area between the graph of f and the x-axis on [a, b] is, 

$$\begin{gathered} a){\int }_0^{5}9-x^{2}dx=\frac{15}{4} \\ b){\int }_0^{5}9-x^{2}dx=15 \end{gathered}$$

Part (a) Step 1. Given information.

We have given,

$f\left(x\right)=9-x^2, [a,b]=\left[0, 5\right], n=5$

Part (a) Step 2. Concept used.

Midpoint Riemann sum formula:

$$\begin{gathered} {\int }_a^{b}f\left(x\right)dx\hspace{0.17em}\approx \Delta \hspace{0.17em}x\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+...+f\left(\frac{x_{n-1}+x_n}{2}\right)\right) \\ \mathrm{where}\hspace{0.17em}\Delta \hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}\frac{b-a}{n} \end{gathered}$$

Lower Riemann sum formula:

${\int }_a^{b}f\left(x\right)dx\hspace{0.17em}\approx \Delta x\left(f\right(x_0)+f(x_1)+f(x_2)+...+f(x_{n-1}\left)\right)$

Part (a) Step 3. Explanation.

We have,

$f\left(x\right)=9-x^2, [a,b]=\left[0, 5\right], n=5$

Length of the subintervals is,

$\Delta \hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}\frac{b-a}{n}=1$

Dividing the interval [0, 5] in to the 5 subinterval with length 1,

$\left[0, 1\right],\left[1, 2\right], \left[2, 3\right],\left[3,4\right],\left[4,5\right]$

So midpoints are:

$\frac{1}{2}, \frac{3}{2},\frac{5}{2},\frac{7}{2},\frac{9}{2}$

Just evaluating the function for those midpoints,

$$\begin{gathered} \frac{1}{2}, \frac{3}{2},\frac{5}{2},\frac{7}{2},\frac{9}{2} \\ f\left(\frac{1}{2}\right)=\frac{35}{4},f\left(\frac{3}{2}\right)=\frac{27}{4}, f\left(\frac{5}{2}\right)=\frac{11}{4},f\left(\frac{7}{2}\right)=-\frac{13}{4},f\left(\frac{9}{2}\right)=-\frac{45}{4} \end{gathered}$$

Using midpoint formula,

$$\begin{gathered} {\int }_0^{5}9-x^{2}dx\approx 1·\left(\frac{35}{4}+\frac{27}{4}+\frac{11}{4}+-\frac{13}{4}+-\frac{45}{4}\right) \\ =\frac{15}{4} \end{gathered}$$

Part (a) Step 4. Conclusion.

Hence, using approximation method to approximate the signed area between the graph of f and the x-axis on [a, b] is,  

${\int }_0^{5}9-x^{2}dx=\frac{15}{4}$

Part (b) Step 1. Explanation.

We have given,

$f\left(x\right)=9-x^2, [a,b]=\left[0, 5\right], n=5$

Length of subintervals is 1 and subintervals are,

$\left[0, 1\right],\left[1, 2\right], \left[2, 3\right],\left[3,4\right],\left[4,5\right]$

Left end points are:

0, 1, 2, 3, 4

Just evaluating the function for those end points,

$f\left(0\right)=9, f\left(1\right)=8, f\left(2\right)=5, f\left(3\right)=0, f\left(4\right)=-7$

Using lower Riemann sum formula,

$$\begin{gathered} {\int }_0^{5}9-x^{2}dx\approx 1·\left(9+8+5+0+-7\right) \\ =15 \end{gathered}$$

Part (b) Step 2. Conclusion.

Hence, using approximation method to approximate the signed area between the graph of f and the x-axis on [a, b] is,  

${\int }_0^{5}9-x^{2}dx=15$