Q. 36 - 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?

$f\left(x\right)=4-x,\left[a,b\right]=\left[0,6\right]$

Step-by-Step Solution

Answer

The actual area of the triangle is 8 square unit.

Area with left sum is 42 square unit.

Step 1. Given Information

The given function is $f\left(x\right)=4-x,\left[a,b\right]=\left[0,6\right]$

The function shows the straight line and makes a triangle with $x$-axis.

Step 2. Finding Area of Triangle

The triangle encloses $28$ full square and $8$ partial squares.

Area of triangle $=28+\frac{1}{2}\left(8\right)=32$ square units.

The actual area of circle is 8 square unit, so the approximation we found is not so approximate.

Step 3. Approximate area with Left Sum

We begin by subdividing interval $\left[0,6\right]$ into four subintervals and $∆x=\frac{b-a}{n}$and $x_k=a+k∆x$.

Here $∆x=\frac{6-0}{n}=\frac{6}{n}$, $x_k=0+k\frac{6}{n}=\frac{6k}{n}$ gives $x_{k-1}=\frac{6\left(k-1\right)}{n}$

$$\begin{gathered} \sum _{k=1}^{n}f\left(x_{k-1}\right)∆x=\sum _{k=1}^{n}f\left(\frac{6\left(k-1\right)}{n}\right)\frac{6}{n} \\ =\sum _{k=1}^4\left[4-\left(\frac{6\left(k-1\right)}{n}\right)\right]\frac{6}{n}=\sum _{k=1}^4\left[\frac{4n-6k+6}{n}\right]\frac{6}{n} \\ ==\sum _{k=1}^4\left[\frac{24n-36k+36}{n}\right] \\ =\left(\frac{24\times 4-36\times 1+36}{4}\right)+\left(\frac{24\times 4-36\times 2+36}{4}\right)+\left(\frac{24\times 4-36\times 3+36}{4}\right)+\left(\frac{24\times 4-36\times 4+36}{4}\right) \\ =24+15+6-3=42 \end{gathered}$$

Step 4. Finding Right Sum

The right sum is given by $\sum _{k=1}^{n}f\left(x_k\right)∆x$

$$\begin{gathered} =\sum _{k=1}^{4}f\left(1.5k\right)1.5=\sum _{k=1}^4\left[4-1.5k\right]1.5= \\ =\left(4-1.5\right)1.5+\left(4-3\right)1.5+\left(4-4.5\right)1.5+\left(4-6\right)1.5 \\ =3.75+1.5-0.75-3=1.5 \end{gathered}$$

Step 5. Finding midpoint sum

$∆x=\frac{b-a}{n}=\frac{6-0}{4}=1.5$

$$\begin{gathered} x_k=a+k∆x=1.5k \\ {x}_{k-1}=1.5\left(k-1\right) \end{gathered}$$

and $f\left(x\right)=4-x$ then $\left(\frac{x_{k-1}+x_k}{2}\right)=\frac{1.5\left(k-1\right)+1.5k}{2}=\frac{3k-1.5}{2}$

$$\begin{gathered} \sum _{k=1}^{n}f\left(\frac{x_{k-1}+x_k}{2}\right)∆x=\sum _{k=1}^{4}f\left(\frac{3k-1.5}{2}\right)1.5 \\ =\sum _{k=1}^4\left(\frac{9.5-3k}{2}\right)1.5=4.8+3.06+0.37-1.8=6.43 \end{gathered}$$

Step 6. Upper Sum

$$\begin{gathered} \sum _{k=1}^{n}f\left(M_k\right)∆x=\sum _{k=1}^{4}f\left(0\right)1.5k \\ =4\times 1.5+4\times 3+4\times 4.5+4\times 6 \\ =60 \end{gathered}$$

Step 7. Finding lower sum

The lower sum is given by $\sum _{k=1}^{n}f\left(m_k\right)∆x$

$$\begin{gathered} =\sum _{k=1}^{4}f\left(m_k\right)∆x=-2\times 1.5+(-2\times 3)+\left(-2\times 4.5\right)+\left(-2\times 6\right) \\ =30 \end{gathered}$$

Step 8. Trapezoid Sum

The trapezoid sum is given by $\sum _{k=1}^n\frac{f\left(x_{k-1}\right)+f\left(x_k\right)}{2}∆x$

$\sum _{k=1}^4\left(\frac{3k-1.5}{2}\right)1.5=18$