# Calculating the area under $2x^2+1$ using the method of exhaustion

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 :)

You have to include $\frac{b}{n}$ in the summation. By adding $\frac{b}{n}$ $n$ times you will end up with $b$.

$$S_n=2\frac{b^3}{n^3}(1^2+2^2+3^2+\cdots+n^2) + n\biggl(\frac{b}{n}\biggr)$$

This can be reduced to:

$$S_n=2\frac{b^3}{n^3}(1^2+2^2+3^2+\cdots+n^2)+b$$

I think you have made an error in calculating the sums. It should be: $$S_1^n = A_1 + A_2 + ... + A_n = \dfrac{2b^{3}}{n^{3}}[1^{2} + 2^{2} + ... + n^{2}] + (\dfrac{b}{n})n$$

Moral of the story: you forget to add up also $$\dfrac{b}{n}$$ n times which would eliminate the denominator n. Same goes for the other sum: $$S_0^n = A_0 + A_2 + ... + A_{n-1} = \dfrac{2b^{3}}{n^{3}}[0^{2} + 1^{2} + 2^{2} + ... + (n-1)^{2}] + (\dfrac{b}{n})n$$

Multiplying the inequality you gave by $$\dfrac{2b^{3}}{n^{3}}$$ and adding up b to it shall give you the answer.

I hope it helps!