# Thread: Another Area Estimate Question

1. ## Another Area Estimate Question

The area between the graph of and the x-axis for 1 ≤ x ≤ 4 s estimated using 20 rectangles and x* k is taken to be the right-hand boundary of each interval.

The result is rounded off to three decimal places.

We devided $b-a=4-1=3$ into 20 rectangles each of width $\frac{3}{20}=0.15$ then the approximation could be done by adding the rectangles $r_k$ of width $\Delta\,x_k\cdot = 0.15$ and length/hight $f(x_k)$ that is $\sum_{k=1}^{20}\,f(x_k)\Delta x_k=4.8912$ whuc will turn out to be very soon to be $\int_1^4\,\sqrt{x}\,dx=\frac{14}{3}=4.66666...$
4. Well if you have reached Rieman sums you should be getting that, but I'm assuming you've never heard about the guy. So, when approximating areas under a curve of $y=f(x)$, above the x-axis and between 2 vertical lines lets say $x=a$ and $x= b$ we divide the segment $b-a$ into partitions of width $\Delta x_k$ each... It's like cutting a piece of bread into slices of equal width (the $\Delta x_k$s don't have to be of equal width though, later to that...). So when you finish dividing the segment into smaller subsegments, and then drawing a recatangle over each subsegment of height = the (average) value of $f$ on that segment (the avergae value of $f$ on $[1;1.15]$ is around 1 because $\sqrt{1} \ \mbox{ almost equal to} \sqrt{1.15} \ \mbox{ almost equal to} 1$)
Then, you add the areas of these rectangles, each has an area of $\ A= \frac{b-a}{20} \cdot f(c)$ where $c$ is any point in the interval $[x_k;x_{k+1}]$ , I would go for the edge points( $\ [x_k;x_{k+1}], \ [1;1.15], \ [1.15;1.30],\ [1.30;1.45]$ and so on...) So when you sum up the whole lot you're doing $\frac{b-a}{20} \cdot ( f(x_0)+f(x_1)+...+f(x_{19}) )$, note that 20 is your choice, it could ran up to 100, a 1000 or infinity.