Answer
(a) $3.56=\frac{89}{25}$
(b) $\frac{11}{3}\approx3.67$
Work Step by Step
(a) Step 1. Divide the region into five equally spaced parts and we can get the x-coordinates when using right endpoints: $x_i=0.2,0.4,0.6,0.8,1$ with a width of $0.2$
Step 2. Using the given equation $f(x)=4-x^2$, we can calculate the heights of the rectangles as $f(x_i)=3.96,3.84,3.64,3.36,3$
Step 3. We can estimate the area by adding up the areas of these rectangles:
$A=\sum 0.2f(x_i)=0.2(3.96+3.84+3.64+3.36+3)=3.56$ or $\frac{356}{100}=\frac{89}{25}$
(b) Step 1. Given the x-interval of $[0,1]$, divide the region into $n$ parts, the width of each part is $\Delta x=\frac{1}{n}$
Step 2. Use the right endpoints, the x-coordinates can be found as $x_k=0+k\Delta x=\frac{k}{n}$
Step 3. Based on the given function $f(x)=4-x^2$, the height of the $k$th rectangle is given by $f(x_k)=4-(\frac{k}{n})^2=4-\frac{k^2}{n^2}$
Step 4. Approximate the area by adding up the areas of all the triangles and evaluate the limit when $n\to\infty$
$A'=\lim_{n\to\infty}\sum^n_{k=1}f(x_k)\Delta x=\lim_{n\to\infty}\sum^n_{k=1}(4-\frac{k^2}{n^2})\frac{1}{n}=\lim_{n\to\infty} ( \frac{1}{n} \sum^n_{k=1} 4-\frac{1}{n^3}\sum^n_{k=1}k^2)=\lim_{n\to\infty} ( \frac{1}{n} \times4n-\frac{1}{n^3}\frac{n(n+1)(2n+1)}{6})=4-\frac{2}{6}=\frac{11}{3}$
Step 5. The area of the region that lies under the graph over the given interval is $A'=\frac{11}{3}\approx3.67$
