Q. 23 - Calculus

Question

For each function $f$ and interval $\left[a,b\right]$ in Exercises $23-25$, use at least eight rectangles to approximate $\left(a\right)$ the signed area and $\left(b\right)$ the absolute area between the graph of $f$ and the $x-axis$ from $x=a$ to$x=b$. Your work should include a graph of $f$ together with the rectangles that you used .

$f\left(x\right)=x^2-3x-4, \left[-5,5\right]$

Step-by-Step Solution

Answer

a. The signed area is $80$.

b. The absolute area is $\approx 43.3334$.

Step 1. Given Information

The function is,

$f\left(x\right)=x^2-3x-4$

The interval is $\left[-5,5\right]$.

Part (a) Step 2. Calculation

The objective is to find the signed area with at least eight rectangles.

The left-sum defined for $n$ rectangles on $\left[a,b\right]$ is $\sum _{k=1}^{n}f\left(x_{k-1}\right)∆x$.

where, $∆x=\frac{b-a}{n}$

               $x_k=a+k∆x$

Now, 

$$\begin{gathered} ∆x=\frac{5+5}{10} \\ =\frac{10}{10} \\ =1 \end{gathered}$$

So,

$$\begin{gathered} x_k=-5+k \\ =k-5 \end{gathered}$$

In the left sum, $x_{k-1}$is the leftmost point in the interval $\left[x_{k-1},x_k\right]$.

So,

$$\begin{gathered} x_{k-1}=k-5-1 \\ =k-6 \end{gathered}$$

The left sum is,

$$\begin{gathered} \sum _{k=1}^{10}\left({\left(k-6\right)}^2-3\left(k-6\right)-4\right)∆x \\ =1\sum _{k=1}^{10}{k}^2-12k+36-3k+18-4 \\ =\sum _{k=1}^{10}{k}^2-15k+50 \\ =\frac{n\left(n+1\right)\left(2n+1\right)}{6}-15\frac{n\left(n+1\right)}{2}+50n \\ =385-825+500 \\ =80 \end{gathered}$$

Therefore, the signed area is $80$.

Part (b) Step 2. Calculation

The objective is to find the absolute area between the graphs of the function and the x-axis

from $x=a$ to $x=b$.

The graph of the function is,

The absolute area is,

$$\begin{gathered} {\int }_{-5}^{5}f\left(x\right)dx \\ ={\int }_{-5}^5{x}^2-3x-4dx \\ ={\int }_{-5}^{1.5}-\left(x^2-3x-4\right)dx+{\int }_{1.5}^5{x}^2-3x-4dx \\ ={\left[\int -\left(x^2-3x-4\right)dx\right]}_{-5}^{1.5}+{\left[\int x^2-3x-4dx\right]}_{1.5}^5 \\ =-{\left[\left(\frac{x^3}{3}\right)-3\left(\frac{x^2}{2}\right)-4x\right]}_{-5}^{1.5}+{\left[\left(\frac{x^3}{3}\right)-3\left(\frac{x^2}{2}\right)-4x\right]}_{1.5}^5 \\ =-\left(\frac{-611}{12}\right)-\frac{91}{12} \\ \approx 50.9167-7.5833 \\ \approx 43.3334 \end{gathered}$$

Therefore, the absolute area is$\approx 43.3334$