Q.27 - 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. 

$f\left(x\right)=x^2, [a,b]=[0,3]$ left sum with

a) n = 3            b) n = 6

Step-by-Step Solution

Answer

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

$$\begin{gathered} a){\int }_0^3{x}^2=5 \\ b){\int }_0^3{x}^2\approx \frac{55}{8} \end{gathered}$$

Part (a) Step 1. Given information.

We have given, 

$f\left(x\right)=x^2 and [a, b] =[0, 3]$

Part (a) Step 2. Concept used.

Left endpoint Riemann sum formula:

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

Where, n is the number of intervals.

Part (a) Step 3. Explanation.

From the given information in step 1. in part (a),

$\Delta \hspace{0.17em}x=\frac{b-a}{3}=\frac{3-0}{3}=1$

Length of the subintervals is 1.

So dividing the interval [0, 3] in to the subintervals with length 1 is,

[0, 1], [1, 2] and [2, 3].

Left end points of those intervals are: 0, 1 and 2.

Now, just evaluating the function at the left endpoints of the subintervals,

$$\begin{gathered} f\left(0\right)=0 \\ f\left(1\right)=1 \\ f\left(2\right)=4 \end{gathered}$$

Using left end point formula,

$$\begin{gathered} {\int }_0^3{x}^2=1[0+1+4] \\ =5 \end{gathered}$$

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,

${\int }_0^3{x}^2=5$

Part (b) Step 1. Explanation

Using given information from part (a) in step 1 and left endpoint formula from the part (a) in step 2.

Using n = 6,

$\Delta \hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}\frac{3-0}{6}=\frac{1}{2}$

So length of interval is $\frac{1}{2}$.

So dividing the interval [0, 3] in to the subintervals with length $\frac{1}{2}$ is,

$\left[0, \frac{1}{2}\right], \left[\frac{1}{2}, 1\right], \left[1, \frac{3}{2}\right], \left[\frac{3}{2}, 2\right], \left[2, \frac{5}{2}\right]and \left[\frac{5}{2}, 3\right]$

Left endpoints are: $0, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}$

Now, just evaluating the function at the left endpoints of the subintervals,

 $f\left(0\right)=0, f\left(\frac{1}{2}\right)=\frac{1}{4}, f\left(\frac{3}{2}\right)=\frac{9}{4}, f\left(2\right)=4, f\left(\frac{5}{2}\right)=\frac{25}{4}$

Using left endpoint formula,

$$\begin{gathered} {\int }_0^3{x}^2\approx 0+\frac{1}{4}+1+\frac{9}{4}+4+\frac{25}{4} \\ {\int }_0^3{x}^2\approx \frac{55}{8} \end{gathered}$$

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,

${\int }_0^3{x}^2\approx \frac{55}{8}$