Q.31 - 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.   

$f\left(x\right)=e^x, \left[a,b\right]=\left[1,4\right], n=6$

(a) midpoint sum                 (b) trapezoid 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,   

$$\begin{gathered} a) {\int }_1^4{e}^x\approx 51.34 \\ b) {\int }_1^4{e}^{x}dx=52.96 \end{gathered}$$

Part (a) Step 1. Given information.

We have given, 

$f\left(x\right)=e^x, \left[a,b\right]=\left[1,4\right], n=6$

Part (a) Step 2. Concept used.

Midpoint formula for Riemann sum:

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

Where,

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

Trapezoidal rule:

${\int }_a^{b}f\left(x\right)dx\hspace{0.17em}\approx \frac{\Delta \hspace{0.17em}x}{2}\left(f\left(x_0\right)+2f\left(x_1\right)+...+2f\left(x_{n-1}\right)+f\left(x_n\right)\right)$

Part (a) Step 3. Explanation.

We have, $f\left(x\right)=e^x, \left[a,b\right]=\left[1,4\right], n=6$

So length of the subintervals is,  

$\hspace{0.17em}\Delta \hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}\frac{b-a}{n}=\frac{1}{2}$

Dividing the interval [1, 4] to 6 subintervals with length $\frac{1}{2}$,

$\left[1, \frac{3}{2}\right],\left[\frac{3}{2}, 2\right],\left[2,\frac{5}{2}\right],\left[\frac{5}{2},3\right],\left[3, \frac{7}{2}\right], \left[\frac{7}{2},4\right]$

So midpoints are: 

$\frac{5}{4}, \frac{7}{4},\frac{9}{4}, \frac{11}{4}, \frac{13}{4},\frac{15}{4}$

Now, just evaluating the function for those midpoints,

$$\begin{gathered} \frac{5}{4}, \frac{7}{4},\frac{9}{4}, \frac{11}{4}, \frac{13}{4},\frac{15}{4} \\ f\left(\frac{5}{4}\right)=e^{\frac{5}{4}},f\left(\frac{7}{4}\right)=e^{\frac{7}{4}} , f\left(\frac{9}{4}\right)=e^{\frac{9}{4}} ,f\left(\frac{11}{4}\right)=e^{\frac{11}{4}} , f\left(\frac{13}{4}\right)=e^{\frac{13}{4}} , f\left(\frac{15}{4}\right)=e^{\frac{15}{4}} \end{gathered}$$

Using midpoint formula,

$$\begin{gathered} {\int }_1^4{e}^x\approx \frac{1}{2}\left(e^{\frac{5}{4}}+e^{\frac{7}{4}}+e^{\frac{9}{4}}+e^{\frac{11}{4}}+e^{\frac{13}{4}}+e^{\frac{15}{4}}\right) \\ \approx 2 1 (3.490342957461841+5.75460267600573+9.487735836358526+15.642631884188172+25.790339917193062+42.521082000062783) \\ =51.343367635635057 \\ \approx 51.34 \end{gathered}$$

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,  

${\int }_1^4{e}^x\approx 51.34$

Part (b) Step 1. Explanation.

We have given,

$f\left(x\right)=e^x, \left[a,b\right]=\left[1,4\right], n=6$

Length of the subintervals is,

$\Delta \hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}\frac{b-a}{n}=\frac{1}{2}$

Dividing the interval [1 , 4] in to the 6 subintervals with length $\frac{1}{2}$,

$\left[1, \frac{3}{2}\right],\left[\frac{3}{2}, 2\right],\left[2,\frac{5}{2}\right],\left[\frac{5}{2},3\right],\left[3, \frac{7}{2}\right], \left[\frac{7}{2},4\right]$

So endpoints are:

$1, \frac{3}{2},2, \frac{5}{2},3,\frac{7}{2}, 4$

Now just evaluating the functions for this endpoints,

$$\begin{gathered} 1, \frac{3}{2},2, \frac{5}{2},3,\frac{7}{2}, 4 \\ f\left(1\right)=e, f\left(\frac{3}{2}\right)=e^{\frac{3}{2}}, f\left(2\right)=e^2, f\left(\frac{5}{2}\right)=e^{\frac{5}{2}}, f\left(3\right)=e^3, f\left(\frac{7}{2}\right)=e^{\frac{7}{2}}, f\left(4\right)=e^4 \end{gathered}$$

Using Trapezoidal rule,

$$\begin{gathered} {\int }_1^4{e}^{x}dx\approx \frac{1}{4}\left(e+2e^{\frac{3}{2}}+2e^2+2e^{\frac{5}{2}}+2e^3+2e^{\frac{7}{2}}+e^4\right) \\ =4 1 (2.718281828459045+8.96337814067613+14.7781121978613+24.364987921406947+40.171073846375335+66.230903917384628+54.598150033144239) \\ =52.956221971326906 \\ \approx 52.96 \end{gathered}$$

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,  

${\int }_1^4{e}^{x}dx=52.96$