Q. 1 - Calculus

Question

Consider the graph of the function f shown next. Define A(x) to be the area of the region between the graph of f and the x-axis from 0 to x. We will count areas of regions above
the x-axis positively and areas of regions below the x-axis negatively.

Use the graph to approximate the values of A(0), A(1), A(2), A(3), A(4), and A(5).

Step-by-Step Solution

Answer

The approximation of the areas of the function of the graph is:

$$\begin{gathered} A\left(0\right)=0 \\ A\left(1\right)=61 \\ A\left(2\right)=152 \\ A\left(3\right)=189 \\ A\left(4\right)=160 \\ A\left(5\right)=125 \end{gathered}$$

Step 1. Given Information.

The graph.

Step 2. Approximate the area.

The area of the function $f\left(x\right)=12x^3-96x^2+180x$

$$\begin{gathered} A\left(x\right)=\frac{12x^4}{4}-\frac{96x^3}{3}+\frac{180x^2}{2} \\ A\left(0\right)=0 \\ A\left(1\right)=\frac{12{\left(1\right)}^4}{4}-\frac{96{\left(1\right)}^3}{3}+\frac{180{\left(1\right)}^2}{2} \\ =61 \\ A\left(2\right)=\frac{12{\left(2\right)}^4}{4}-\frac{96{\left(2\right)}^3}{3}+\frac{180{\left(2\right)}^2}{2} \\ =152 \\ A\left(3\right)=\frac{12{\left(3\right)}^4}{4}-\frac{96{\left(3\right)}^3}{3}+\frac{180{\left(3\right)}^2}{2} \\ =189 \\ A\left(4\right)=\frac{12{\left(4\right)}^4}{4}-\frac{96{\left(4\right)}^3}{3}+\frac{180{\left(4\right)}^2}{2} \\ =160 \\ A\left(5\right)=\frac{12{\left(5\right)}^4}{4}-\frac{96{\left(5\right)}^3}{3}+\frac{180{\left(5\right)}^2}{2} \\ =125 \end{gathered}$$