I'm currently reading Tom M Apostol Calculus Volume I, and I'm stuck on part b) of the following question, regarding the method of exhaustion in calculating the area under a curve:
Modify the region in the figure below by assuming the ordinate at each $x$ is a) $2x^2$ b)$2x^2+1$ instead of $x^2$. Check through the principal steps, and find what effect this has on the calculation of the area.
I will quickly run through what information is available, using the calculation of the area under the curve $x^2$ (please excuse my quickly drawn diagrams)
We are calculating the area under the curve from $0$ to $b$. The base of the curve is subdivided into $n$ equal parts, each of length $\frac bn$. A typical point of subdivision corresponds to $x=\frac {kb}{n}$, where $k$ takes the successive values $k=0,1,2,3,...,n$
The area of typical outer rectangle $A_O$, is clearly: $$A_O = \left( \frac bn \right) \left( \frac{kb}{n} \right) ^2 = \frac{b^3}{n^3} k^2$$
Similarly, the area of a typical inner rectangle is:
$$A_O = \left( \frac bn \right) \left( \frac{(k-1)b}{n} \right) ^2 = \frac{b^3}{n^3} (k-1)^2$$
The sum of all the outer rectangles $S_n^O$ is: $$S_n^O = \frac {b^3}{n^3}[1^2+2^2+3^2+...+n^2] \ \mathrm{(eq.1)}$$
The sum of all the inner rectangles $S_n^I$ is: $$S_n^I = \frac {b^3}{n^3}[1^2+2^2+3^2+...+(n-1)^2] \ \mathrm{(eq.2)}$$
We are given the following inequalities, which were proved earlier in the book:
$$1^2+2^2+...+(n-1)^2 \lt \frac {n^3}{3} \lt 1^2+2^2+...+n^2 \ \mathrm{for} \ n \ge 1 \ \mathrm{(eq.3)}$$
Now, multiplying by $\frac{b^3}{n^3}$ then simplifying and substituting eq.1 and eq.2 gives:
$$S_n^I \lt \frac{b^3}{3} \lt S_n^O$$
Part a) of the question is straight forward, and I will not redraw the figure as it is almost identical.
a) $2x^2$
Running through the steps above with the ordinate set as $2x^2$ is as follows:
The area of typical outer rectangle $A_O$, is clearly: $$A_O = 2\left( \frac bn \right) \left( \frac{kb}{n} \right) ^2 = 2\frac{b^3}{n^3} k^2$$
Similarly, the area of a typical inner rectangle is:
$$A_I = 2\left( \frac bn \right) \left( \frac{(k-1)b}{n} \right) ^2 = 2\frac{b^3}{n^3} (k-1)^2$$
The sum of all the outer rectangles $S_n^O$ is: $$S_n^O = \frac {2b^3}{n^3}[1^2+2^2+3^2+...+n^2] \ \mathrm{(eq.4)}$$
The sum of all the inner rectangles $S_n^I$ is: $$S_n^I = \frac {2b^3}{n^3}[1^2+2^2+3^2+...+(n-1)^2] \ \mathrm{(eq.5)}$$
We can again use the following inequalities:
$$1^2+2^2+...+(n-1)^2 \lt \frac {n^3}{3} \lt 1^2+2^2+...+n^2 \ \mathrm{for} \ n \ge 1 \ \mathrm{(eq.3)}$$
Now, multiplying by $\frac{2b^3}{n^3}$ then simplifying and substituting eq.3 and eq.4 gives:
$$S_n^I \lt \frac{2b^3}{3} \lt S_n^O$$
Okay, so far so good. The above result is as expected.
Now, for part b). I will go through my workings and show you where I get stuck.
b)$2x^2+1$
Below, is the redrawn figures using $2x^2+1$ as the ordinate:
I've calculated the areas of the rectangles as follows:
$$A_O = \left( \frac bn \right) \left( 2\left( \frac{kb}{n} \right) ^2 + 1 \right) = \frac bn + \frac{2b^3}{n^3} k^2$$
$$A_I = \left( \frac bn \right) \left( 2\left( \frac{(k-1)b}{n} \right) ^2 + 1 \right) = \frac bn + \frac{2b^3}{n^3} (k-1)^2$$
The area of the sum of all outer rectangles I've calculated as:
$$S_n^I = \frac bn + \frac{2b^3}{n^3} [1^2+2^2+3^2+...+n^2]$$
The area of the inner rectangles I've calculated as: (Note, the extra rectangle from 0 to $b/n$)
$$S_n^O = \frac {2b}{n} + \frac{2b^3}{n^3} [1^2+2^2+3^2+...+(n-1)^2]$$
Again, we can use the following inequalities:
$$1^2+2^2+...+(n-1)^2 \lt \frac {n^3}{3} \lt 1^2+2^2+...+n^2 \ \mathrm{for} \ n \ge 1 \ \mathrm{(eq.3)}$$
Now this is where I get stuck. If I multiply the above by $\frac {2b^3}{3}$ I get:
$$S_n^I - \frac {2b}{n} \lt \frac{2b^3}{3} \lt S_n^O - \frac bn$$
From here I can't understand how I can possibly eliminate the $n$'s and come to the desired result:
$$S_n^I \lt \frac{2b^3}{3} + b \lt S_n^O$$
Any help would be MUCH appreciated :)
