In Problems 41 and 42, a function f is defined over an interval [a, b]

(a) Graph f, indicating the area A under f from a to b.

(b) Approximate the area A by partitioning [a, b] into three subintervals of equal length and choosing u as the left endpoint of each subinterval.

(c) Approximate the area A by partitioning [a, b] into six subintervals of equal length and choosing u as the left endpoint of each subinterval.

(d) Express area A as an integral.

(e) Use a graphing utility to approximate the integral.

$f\left(x\right)=4-x^2,[-1,2]$

Consider the given information,

By using a graphing utility,

The graph of function including area is shown below,

Subinterval length $(∆x)$is computed as,

$$\begin{gathered} ∆x=\frac{b-a}{n} \\ ∆x=\frac{2-(-1)}{3}=\frac{3}{3}=1 \end{gathered}$$

So, three sub-intervals are,

The area is computed as;

$$\begin{gathered} A=\left[f\right(-1)+f(0)+f(1\left)\right]\times 1 \\ A=[3+4+3]\times 1 \\ A=10\times 1 \\ A=10 \end{gathered}$$

Subinterval length is computed as,

$\begin{array}{rcl}\Delta x& =& \frac{2-\left(-1\right)}{8}\\ & =& \frac{3}{8}=0.375\end{array}$

So, eight sub-intervals are,

role="math" localid="1652764241943" $$\begin{gathered} \left[-1,-0.625\right],\left[-0.625,-0.25\right],\left[-0.25,0.125\right],\left[0.125,0.5\right], \\ \left[0.5,0.875\right],\left[0.875,1.25\right],\left[1.25,1.625\right],\left[1.625,2\right] \end{gathered}$$

The area is computed as,

$\begin{array}{rcl}A& \approx & 0.375\left[f\left(-1\right)+f\left(-0.625\right)+f\left(-0.25\right)+f\left(0.125\right)+f\left(0.5\right)+f\left(0.875\right)+f\left(1.25\right)+f\left(1.625\right)\right]\\ & =& 0.375\left[\left(4-{\left(-1\right)}^2\right)+\left(4-{\left(-0.625\right)}^2\right)+\left(4-{\left(-0.25\right)}^2\right)+\left(4-{\left(0.125\right)}^2\right)+\left(4-{\left(0.5\right)}^2\right)+\left(4-{\left(0.875\right)}^2\right)+\left(4-{\left(1.25\right)}^2\right)+\left(4-{\left(1.625\right)}^2\right)\right]\\ & \approx & 9.4922\end{array}$

Consider the given function and integral.

The area is defined as,

$A={\int }_a^{b}f\left(x\right)dx$

Substitute the values,

$A={\int }_{-1}^2\left(4-x^2\right)dx$

Use the calculator to find the integral.

$\begin{array}{rcl}A& =& {\int }_{-1}^2\left(4-x^2\right)dx\\ & =& 9\end{array}$