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Use the method of exhaustion to calculate the area under the following curves.

Calculate the areas under the following curves using the method of exhaustion:

  1. 2x^2;
  2. 3x^2;
  3. \frac{1}{4}x^2;
  4. 2x^2 + 1;
  5. ax^2 + c.

  1. 2x^2.
    The following graphs show the lower and upper sums, respectively:

    Rendered by QuickLaTeX.com

    Rendered by QuickLaTeX.com

    So, now we need to compute the lower and upper sums. For the lower sums we add up the area of all of the blue rectangle. Each rectangle has width \frac{b}{n} (since we have divided the interval [0,b] into n equal pieces) and its height is the value of the curve on the left side of the rectangle (since the curve is drawn so that the left corner rests on the curve). Therefore, we have

        \[ s_n = \frac{b}{n} \cdot  2 \left( \frac{b}{n} \right)^2 +  \frac{b}{n} \cdot  2 \left( \frac{2b}{n} \right)^2 + \cdots +  \frac{b}{n} \cdot  2 \left( \frac{(n-1)b}{n} \right)^2. \]

    For the upper sums the width of each rectangle is still \frac{b}{n}, but the height is now given by the function value on the right corner. So we have,

        \[ S_n = \frac{b}{n} \cdot 2 \left( \frac{b}{n} \right)^2 + \frac{b}{n} \cdot 2 \left( \frac{2b}{n} \right)^2 + \cdots + \frac{b}{n} \cdot 2 \left( \frac{(n-1)b}{n} \right)^2 + \frac{b}{n} \cdot 2 \left( \frac{bn}{n} \right)^2. \]

    Now, we are given the identity

        \[ \sum_{i=1}^n i^2 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}, \]

    and so, following the example in Apostol,

        \[ 1^2 + 2^2 + \cdots + (n-1)^2 < \frac{n^3}{3} < 1^2 + 2^2 + \cdots + n^2. \]

    Now, we multiply each term in the inequality by \frac{b}{n} \cdot 2 \left( \frac{b}{n} \right)^2 so the sums on the far left and far right become our s_n and S_n above, and we can simplify the term in the middle. Thus, for all n \geq 1 we have

        \begin{align*}   s_n &< \frac{n^3}{3} \cdot \frac{b}{n} \cdot 2 \left(\frac{b}{n} \right)^2 < S_n \\ \implies s_n &< \frac{2b^3}{3} < S_n. \end{align*}

    Now, we must show that if s_n < A < S_n for all n \geq 1 then A = \frac{2b^3}{3}. We accomplish this by showing that A < \frac{2b^3}{3} and A > \frac{2b^3}{3} both lead to contradictions.

    Suppose A < \frac{2b^3}{3}. Then, since S_n - s_n = \frac{2b^3}{n} we can compute,

        \begin{align*}  \frac{2b^3}{3} < S_n && \implies && \frac{2b^3}{3} - s_n &< S_n - s_n \\  && \implies && \frac{2b^3}{3} - A &< \frac{2b^3}{n} \\  && \implies && n &< \frac{2b^3}{\left( \frac{2b^3}{3} \right) - A}. \end{align*}

    However, this cannot hold for all n \geq 1 since the term on the right is a constant (depending on b), so we can always choose some n that is larger since the positive integers are unbounded above.

    Next, suppose A > \frac{2b^3}{3}. Then,

        \begin{align*}  A < S_n && \implies && A - s_n &< S_n - s_n \\  && \implies && A - \frac{2b^3}{3} &< \frac{2b^3}{n} \\  && \implies && n &< \frac{2b^3}{A - \left( \frac{2b^3}{3} \right)}. \end{align*}

    But, again, this cannot be true for all positive integers n since the positive integers are unbounded above. Therefore, both A < \frac{2b^3}{3} and A > \frac{2b^3}{3} lead to contradictions. Hence, we must have

        \[ A = \frac{2b^3}{3}. \qquad \blacksquare \]

  2. 3x^2.
    The following graphs show the lower and upper sums, respectively:

    Rendered by QuickLaTeX.com

    Rendered by QuickLaTeX.com

    As in part (a), we need to compute the lower and upper sums. For the lower sums we add up the area of all of the blue rectangle. Each rectangle has width \frac{b}{n} (since we have divided the interval [0,b] into n equal pieces) and its height is the value of the curve on the left side of the rectangle (since the curve is drawn so that the left corner rests on the curve). Therefore, we have

        \[ s_n = \frac{b}{n} \cdot  3 \left( \frac{b}{n} \right)^2 +  \frac{b}{n} \cdot  3 \left( \frac{2b}{n} \right)^2 + \cdots +  \frac{b}{n} \cdot  3 \left( \frac{(n-1)b}{n} \right)^2. \]

    For the upper sums the width of each rectangle is still \frac{b}{n}, but the height is now given by the function value on the right corner. So we have,

        \[ S_n = \frac{b}{n} \cdot 3 \left( \frac{b}{n} \right)^2 + \frac{b}{n} \cdot 3 \left( \frac{2b}{n} \right)^2 + \cdots + \frac{b}{n} \cdot 3 \left( \frac{(n-1)b}{n} \right)^2 + \frac{b}{n} \cdot 3 \left( \frac{bn}{n} \right)^2. \]

    Now, we are given the identity

        \[ \sum_{i=1}^n i^2 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}, \]

    and so, following the example in Apostol,

        \[ 1^2 + 2^2 + \cdots + (n-1)^2 < \frac{n^3}{3} < 1^2 + 2^2 + \cdots + n^2. \]

    Now, we multiply each term in the inequality by \frac{b}{n} \cdot 3 \left( \frac{b}{n} \right)^2 so the sums on the far left and far right become our s_n and S_n above, and we can simplify the term in the middle. Thus, for all n \geq 1 we have

        \begin{align*}   s_n &< \frac{n^3}{3} \cdot \frac{b}{n} \cdot 3 \left(\frac{b}{n} \right)^2 < S_n \\ \implies s_n &< b^3 < S_n. \end{align*}

    Now, we must show that if s_n < A < S_n for all n \geq 1 then A = b^3. We accomplish this by showing that A < b^3 and A > b^3 both lead to contradictions.

    Suppose A < b^3. Then, since S_n - s_n = \frac{3b^3}{n} we can compute,

        \begin{align*}  b^3 < S_n && \implies && b^3 - s_n &< S_n - s_n \\  && \implies && b^3 - A &< \frac{3b^3}{n} \\  && \implies && n &< \frac{3b^3}{b^3 - A}. \end{align*}

    However, this cannot hold for all n \geq 1 since the term on the right is a constant (depending on b), so we can always choose some n that is larger since the positive integers are unbounded above.

    Next, suppose A > b^3. Then,

        \begin{align*}  A < S_n && \implies && A - s_n &< S_n - s_n \\  && \implies && A - b^3 &< \frac{3b^3}{n} \\  && \implies && n &< \frac{3b^3}{A - b^3}. \end{align*}

    But, again, this cannot be true for all positive integers n since the positive integers are unbounded above. Therefore, both A < b^3 and A > b^3 lead to contradictions. Hence, we must have

        \[ A = b^3. \qquad \blacksquare \]

  3. \frac{1}{4} x^2
    The following graphs show the lower and upper sums, respectively:

    Rendered by QuickLaTeX.com

    Rendered by QuickLaTeX.com

    This will be very similar to parts (a) and (b) except we have the factor of \frac{1}{4} instead of 2 or 3. First, we need to compute the lower and upper sums. For the lower sums we add up the area of all of the blue rectangle. Each rectangle has width \frac{b}{n} (since we have divided the interval [0,b] into n equal pieces) and its height is the value of the curve on the left side of the rectangle (since the curve is drawn so that the left corner rests on the curve). Therefore, we have

        \[ s_n = \frac{b}{n} \cdot  \frac{1}{4} \left( \frac{b}{n} \right)^2 +  \frac{b}{n} \cdot  \frac{1}{4} \left( \frac{2b}{n} \right)^2 + \cdots +  \frac{b}{n} \cdot  \frac{1}{4} \left( \frac{(n-1)b}{n} \right)^2. \]

    For the upper sums the width of each rectangle is still \frac{b}{n}, but the height is now given by the function value on the right corner. So we have,

        \[ S_n = \frac{b}{n} \cdot \frac{1}{4} \left( \frac{b}{n} \right)^2 + \frac{b}{n} \cdot \frac{1}{4} \left( \frac{2b}{n} \right)^2 + \cdots + \frac{b}{n} \cdot \frac{1}{4} \left( \frac{(n-1)b}{n} \right)^2 + \frac{b}{n} \cdot \frac{1}{4} \left( \frac{bn}{n} \right)^2. \]

    Now, we are given the identity

        \[ \sum_{i=1}^n i^2 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}, \]

    and so, following the example in Apostol,

        \[ 1^2 + 2^2 + \cdots + (n-1)^2 < \frac{n^3}{3} < 1^2 + 2^2 + \cdots + n^2. \]

    Now, we multiply each term in the inequality by \frac{b}{n} \cdot \frac{1}{4} \left( \frac{b}{n} \right)^2 so the sums on the far left and far right become our s_n and S_n above, and we can simplify the term in the middle. Thus, for all n \geq 1 we have

        \begin{align*}   s_n &< \frac{n^3}{3} \cdot \frac{b}{n} \cdot \frac{1}{4} \left(\frac{b}{n} \right)^2 < S_n \\ \implies s_n &< \frac{b^3}{12} < S_n. \end{align*}

    Now, we must show that if s_n < A < S_n for all n \geq 1 then A = \frac{b^3}{12}. We accomplish this by showing that A < \frac{b^3}{12} and A > \frac{b^3}{12} both lead to contradictions.

    Suppose A < \frac{b^3}{12}. Then, since S_n - s_n = \frac{b^3}{4n} we can compute,

        \begin{align*}  \frac{b^3}{12} < S_n && \implies && \frac{b^3}{12} - s_n &< S_n - s_n \\  && \implies && \frac{b^3}{12} - A &< \frac{b^3}{4n} \\  && \implies && n &< \frac{b^3}{4 \left( \frac{b^3}{3}  - A\right)}. \end{align*}

    However, this cannot hold for all n \geq 1 since the term on the right is a constant (depending on b), so we can always choose some n that is larger since the positive integers are unbounded above.

    Next, suppose A > \frac{b^3}{12}. Then,

        \begin{align*}  A < S_n && \implies && A - s_n &< S_n - s_n \\  && \implies && A - \frac{b^3}{12} &< \frac{b^3}{4n} \\  && \implies && n &< \frac{b^3}{4 \left(A - \frac{b^3}{12} \right)}. \end{align*}

    But, again, this cannot be true for all positive integers n since the positive integers are unbounded above. Therefore, both A < \frac{b^3}{12} and A > \frac{b^3}{12} lead to contradictions. Hence, we must have

        \[ A = \frac{b^3}{12}. \qquad \blacksquare \]

  4. 2x^2+1.
    The following graphs show the lower and upper sums, respectively:

    Rendered by QuickLaTeX.com

    Rendered by QuickLaTeX.com

    Again, we need to compute the lower and upper sums. For the lower sums we add up the area of all of the blue rectangle. Each rectangle has width \frac{b}{n} (since we have divided the interval [0,b] into n equal pieces) and its height is the value of the curve on the left side of the rectangle (since the curve is drawn so that the left corner rests on the curve). Therefore, we have

        \[ s_n = \frac{b}{n} \cdot  \left(2 \left( \frac{b}{n} \right)^2 + 1\right) +  \frac{b}{n} \cdot  \left(2 \left( \frac{2b}{n} \right)^2 + 1 \right) + \cdots +  \frac{b}{n} \cdot  \left(2 \left( \frac{(n-1)b}{n} \right)^2 + 1\right). \]

    For the upper sums the width of each rectangle is still \frac{b}{n}, but the height is now given by the function value on the right corner. So we have,

        \[ S_n = \frac{b}{n} \cdot \left(2 \left( \frac{b}{n} \right)^2 + 1 \right) + \frac{b}{n} \cdot \left(2 \left( \frac{2b}{n} \right)^2 + 1 \right)+ \cdots + \frac{b}{n} \cdot \left(2 \left( \frac{(n-1)b}{n} \right)^2 + 1\right) + \frac{b}{n} \cdot \left( 2 \left( \frac{bn}{n} \right)^2 + 1 \right). \]

    Now, we are given the identity

        \[ \sum_{i=1}^n i^2 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}, \]

    and so, following the example in Apostol,

        \[ 1^2 + 2^2 + \cdots + (n-1)^2 < \frac{n^3}{3} < 1^2 + 2^2 + \cdots + n^2. \]

    Next, to get the inequalities we want (with s_n < \cdots < S_n) it takes a bit more work than in parts (a) – (c) since we now have the +1 term to deal with. First, we multiply each term in the inequality by \frac{b}{n} \cdot 2 \left( \frac{b}{n} \right)^2 to obtain

        \begin{alignat*}{3}  && \ 2 \cdot \left( \frac{b}{n} \right)^3 (1^2 + \cdots + (n-1)^2) & \ <\ & \frac{2b^3}{3} \phantom{+n} & \ < \  2 \cdot \left( \frac{b}{n} \right)^3 (1^2 + \cdots + n^2) \\ \implies && \ 2 \cdot \left( \frac{b}{n} \right)^3 (1^2 + \cdots + (n-1)^2) + \frac{nb}{n} & \ <\ & \frac{2b^3}{3} + \frac{nb}{n} & \ < \  2 \cdot \left( \frac{b}{n} \right)^3 (1^2 + \cdots + n^2) + \frac{nb}{n}\\ \implies && \ 2 \cdot \left( \frac{b}{n} \right)^3 (1^2 + \cdots + (n-1)^2) + \frac{b(n-1)}{n} & \ <\ & \frac{2b^3}{3} + b & \ < \  2 \cdot \left( \frac{b}{n} \right)^3 (1^2 + \cdots + n^2) + \frac{nb}{n}\\ \implies && \ s_n & \ < \ & \frac{2b^3}{3} + b & \ < \ S_n. \end{alignat*} \end{align*}

    Now, we must show that if s_n < A < S_n for all n \geq 1 then A = \frac{2b^3}{3} + b. We accomplish this by showing that A < \frac{2b^3}{3} + b and A > \frac{2b^3}{3} + b both lead to contradictions.

    Suppose A < \frac{2b^3}{3} + b. Then, since S_n - s_n = \frac{2b^3}{n} + \frac{b}{n} we can compute,

        \begin{align*}  \frac{2b^3}{3} + b < S_n && \implies && \frac{2b^3}{3} + b - s_n &< S_n - s_n \\  && \implies && \frac{2b^3}{3} + b - A &< \frac{2b^3}{n} + \frac{b}{n} \\  && \implies && n &< \frac{2b^3 + b}{\left( \frac{2b^3}{3}  + b\right) - A}. \end{align*}

    However, this cannot hold for all n \geq 1 since the term on the right is a constant (depending on b), so we can always choose some n that is larger since the positive integers are unbounded above.

    Next, suppose A > \frac{2b^3}{3} + b. Then,

        \begin{align*}  A < S_n && \implies && A - s_n &< S_n - s_n \\  && \implies && A - \frac{2b^3}{3} - b &< \frac{2b^3}{n} + b\\  && \implies && n &< \frac{2b^3+ b}{A - \left( \frac{2b^3}{3}  + b\right)}. \end{align*}

    But, again, this cannot be true for all positive integers n since the positive integers are unbounded above. Therefore, both A < \frac{2b^3}{3} + b and A > \frac{2b^3}{3} + b lead to contradictions. Hence, we must have

        \[ A = \frac{2b^3}{3} + b. \qquad \blacksquare \]

  5. ax^2 + c.

    Finally, we follow the same methods as in parts (a) – (d) to compute the lower and upper sums,

        \begin{align*}  s_n &= \frac{b}{n}  \left( a \left(\frac{b}{n} \right)^2 + c \right) +  \frac{b}{n}  \left( a \cdot \left(\frac{2b}{n}\right)^2 + c \right) + \cdots + \frac{b}{n}  \left( a \cdot \left( \frac{b(n-1)}{n} \right)^2 + c \right) \\[9pt]  &= a \left( \frac{b}{n} \right)^3 \left(1^2 + \cdots + (n-1)^2\right) + \left( \frac{b}{n} \right) \left( (n-1) c \right) \\[9pt] S_n &= \frac{b}{n}  \left( a \left( \frac{b}{n} \right)^2 + c \right) + \frac{b}{n} \left( a \left( \frac{2b}{n} \right)^2 + c \right) + \cdots + \frac{b}{n} \left( a \left( \frac{nb}{n} \right)^2 + c \right) \\[9pt]  &= a \left( \frac{b}{n} \right)^3 (1^2 + \cdots + n^2) + \left( \frac{b}{n} \right) (nc). \end{align*}

    Then, similar to part (d), we have

        \[ a \left( \frac{b}{n} \right)^3 (1^2+ \cdots + (n-1)^2) + bc - \frac{bc}{n} < \frac{ab^3}{3} + bc < a \left( \frac{b}{n} \right)^3 (1^2 + \cdots + n^2) + bc. \]

    Therefore,

        \[ s_n < \frac{ab^3}{3} + bc < S_n. \]

    Now, we must show that if s_n < A < S_n for all n \geq 1 then A = \frac{ab^3}{3} + bc. We accomplish this by showing that A < \frac{ab^3}{3} + bc and A > \frac{ab^3}{3} + bc both lead to contradictions.

    Suppose A < \frac{ab^3}{3} + bc. Then, since S_n - s_n = \frac{ab^3}{n} + \frac{bc}{n} we can compute,

        \begin{align*}  \frac{ab^3}{3} + bc < S_n && \implies && \frac{ab^3}{3} + bc - s_n &< S_n - s_n \\  && \implies && \frac{ab^3}{3} + bc - A &< \frac{ab^3}{n} + \frac{bc}{n} \\  && \implies && n &< \frac{ab^3 + bc}{\left( \frac{ab^3}{3}  + bc\right) - A}. \end{align*}

    However, this cannot hold for all n \geq 1 since the term on the right is a constant (depending on b), so we can always choose some n that is larger since the positive integers are unbounded above.

    Next, suppose A > \frac{ab^3}{3} + bc. Then,

        \begin{align*}  A < S_n && \implies && A - s_n &< S_n - s_n \\  && \implies && A - \frac{ab^3}{3} - bc &< \frac{ab^3}{n} + bc\\  && \implies && n &< \frac{ab^3+ bc}{A - \left( \frac{ab^3}{3}  + bc\right)}. \end{align*}

    But, again, this cannot be true for all positive integers n since the positive integers are unbounded above. Therefore, both A < \frac{ab^3}{3} + bc and A > \frac{ab^3}{3} + bc lead to contradictions. Hence, we must have

        \[ A = \frac{ab^3}{3} + bc. \qquad \blacksquare \]

7 comments

  1. Nayla says:

    Hi,

    In the exercise 1.d and 1.e when you get to the inequalities (that end up with :
    sn< (2(b`3)/3)+b)<Sn) first, you multiply each term in the inequality by: (b/n. 2.(b/n)`2) and then you add (nb/n). So far it's ok , but in the third step on the left side we have ((n-1).b/n) instead of (nb/n) that we add in the previous step. what is the arithmetic done there that allows you to write that?. Of course we need that to get the area of the inner sums, but in terms of arithmetic i don't know how you back that up without affecting the left side of the inequality. Im not sure how can you get to that result without affecting the right side of the inequality. I tried to treat them separately and i have two different area assumptions that don't allow me to put them in the same inequality because when i try to build the inner area sums sn i have to add (- b/n ) to achieve
    ((n-1).b/n).

    I hope it's clear enough,

    Thank you

  2. wells says:

    Hi, I also want to type up my own solution but I had some troubles with the graph. Are you using the pfgplots for these graphs? I am new to latex and cannot figure out how to graph.. Any helps will be appreciated.

  3. JpB says:

    I think I found a mistake, in d) You state Sn-sn=(2b^3/n)+(b/n) but I get Sn-sn=(2b^3/n). Both yield the same overall result showing there is a contradiction with the unbounded n. Am I wrong or did I spot something? Because if I did I will feel pretty cool.

  4. Subhankar says:

    Hi Rori! Thanks for the great work! I have a question regarding Part (d). In the inequalities that you have written following the statement: “Next, to get the inequalities we want (with s_n < \cdots < S_n) it takes a bit more work than in parts (a) – (c) since we now have the +1 term to deal with. First, we multiply each term in the inequality", there seems to be a slight inconsistency. Inequality 2 arises as you have added nb/n throughout. However, in Inequality 3, for the leftmost identity, you change nb/n to (n-1)b/n. Now, the leftmost identity should have (n-1)b/n since there are n-1 terms for the leftmost identity. Nevertheless, the result that you have in Inequality 3 is not implied from Inequality 2 unless you can show nb/n = 0, such that you can drop that term and you end up (n-1)b/n as the last term in your leftmost identity. But this is not the case.

    I hope you understood my issue! Specifically, I am wondering, for the leftmost identity, how did you end up with Inequality 3 from Inequality 2. How or why did you drop the nb/n term?

    Thank you so much!

      • Gustavo Rodríguez says:

        I think you were right but the problem is the definition of Sn and sn.
        For Sn you must take the “right side of the rectangle” (the height is given by the end of the interval for f(x)) so your addition goes from (b/n) to b, but for sn you must take the “left side” (the height is given by the begining of the interval for f(x)) hence the addition goes from 0 to b-b/n. Both additions have n terms (just different heights) but if c=0 the first rectangle of sn cancels out (height=0), else you get height=c and it adds a term to the sum. Therefore the last term of Sn and sn should be nb/n instead of (n-1)b/n for the latter.

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