Q. 38 - Calculus

Question

For each function f and interval [a, b] in Exercises 34–38, it is possible to find the exact signed area between the graph of f and the x-axis on [a, b] geometrically by using the areas of circles,

triangles, and rectangles. Find this exact area, and then calculate the left, right, midpoint, upper, lower, and trapezoid sums with n = 4. Which approximation rule is most accurate?

$f\left(x\right)=3+\sqrt{4-x^2},\left[a,b\right]=\left[-2,2\right]$

Step-by-Step Solution

Answer

The actual area of the semicircle is 21 square units

Step 1. Given Information

The given function is $f\left(x\right)=3+\sqrt{4-x^2},\left[a,b\right]=\left[-2,2\right]$

The function shows semicircle 3 units above from origin

Step 2. Finding area of semicircle

The semicircle has 16 full square and 10 partial square.

So, area is$16+\frac{1}{2}\times 10=21$ square units.

Step 2. Left Sum

$$\begin{gathered} ∆x=\frac{b-a}{n}=\frac{2+2}{4}=1 \\ {x}_k=a+k∆x=-2+k \\ {x}_{k-1}=-2+k-1=-3+k \end{gathered}$$

$f\left(x_{k-1}\right)=3+\sqrt{4-{\left(k-3\right)}^2}$

The left sum$=\sum _{k=1}^{n}f\left(x_{k-1}\right)∆x$

$$\begin{gathered} =\sum _{k=1}^{4}3+\sqrt{4-{\left(k-3\right)}^2} \\ =17.46 \end{gathered}$$

Step 4. Right Sum

The Right sum is

$=\sum _{k=1}^{n}f\left(x_k\right)∆x$

$$\begin{gathered} =\sum _{k=1}^{4}3+\sqrt{4-{\left(k-2\right)}^2} \\ =17.46 \end{gathered}$$

Step 5. Midpoint sum

$\frac{x_{k-1}+x_k}{2}=\frac{k-2+k-3}{2}=\frac{2k-5}{2}$

$f\left(\frac{2k-5}{2}\right)=3+\sqrt{4-{\left(\frac{2k-5}{2}\right)}^2}$

The midpoint sum is 

$$\begin{gathered} =\sum _{k=1}^{4}f\left(\frac{2k-5}{2}\right)∆x=\sum _{k=1}^{4}3+\sqrt{4-{\left(\frac{2k-5}{2}\right)}^2} \\ =16.88 \end{gathered}$$

Step 6. Upper sum

The upper sum is

 $\sum _{k=1}^{n}f\left(M_k\right)∆x$

$$\begin{gathered} f\left(-1\right)∆x+f\left(0\right)∆x+f\left(1\right)∆x+f\left(2\right)∆x \\ =4.73+5+4.73+3=17.46 \end{gathered}$$

Step 7. Lower Sum

The lower sum is

 $\sum _{k=1}^{n}f\left(m_k\right)∆x$

$$\begin{gathered} f\left(-2\right)∆x+f\left(-1\right)∆x+f\left(0\right)∆x+f\left(1\right)∆x \\ =3+4.73+5+4.73=17.46 \end{gathered}$$

Step 8. Trapezoid Sum

$\frac{f\left(x_{k-1}\right)+f\left(x_k\right)}{2}=\frac{3+\sqrt{4-{\left(k-3\right)}^2}+3+\sqrt{4-{\left(k-2\right)}^2}}{2}$

The midpoint sum is

 $$\begin{gathered} \sum _{k=1}^4\frac{3+\sqrt{4-{\left(k-3\right)}^2}+3+\sqrt{4-{\left(k-2\right)}^2}}{2}∆x \\ =2\left(7+\sqrt{3}\right) \end{gathered}$$