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Use the method of exhaustion to calculate the area under more general curves

by RoRi

Modify Figure I.3 so that the ordinate at each point is x^3 instead of x^2.

  1. Show that the outer and inner sums are given by

        \[ S_n = \frac{b^4}{n^4} (1^3+ 2^3 + \cdots +n^3), \qquad s_n = \frac{b^4}{n^4}\left( 1^3 + 2^3 + \cdots +(n-1)^3 \right) \]

  2. Use the inequalities

        \[ 1^3+2^3 + \cdots +(n-1)^3 < \frac{n^4}{4} < 1^3+2^3 + \cdots +n^3 \]

    to show s_n < \frac{b^4}{4} < S_n for all n, and prove that \frac{b^4}{4} is the only number with this property.

  3. What number is in place of \frac{b^4}{4} if the ordinates at each point are given by ax^3+c?
  1. Similar to the previous exercise we compute the sums s_n and S_n by summing up rectangles lying below and above the curve x^3, respectively. Each rectangle has width \frac{b}{n} since we are dividing the interval [0,b] into n equal segments. For s_n the height of each rectangle is the value of x^3 on the left edge of the interval, and for S_n the height of each rectangle is the value of x^3 on the right edge of the interval. Therefore, we have

        \begin{align*} s_n &= \frac{b}{n} \left( \frac{b}{n} \right)^3 + \frac{b}{n} \left( \frac{2b}{n} \right)^3 + \cdots + \frac{b}{n} \left( \frac{(n-1)b}{n} \right)^3 \\[9pt] &= \left( \frac{b}{n} \right)^4 (1^3 + 2^3 + \cdots + (n-1)^3). \end{align*}

    and,

        \begin{align*} S_n &= \frac{b}{n} \left( \frac{b}{n} \right)^3 + \frac{b}{n} \left( \frac{2b}{n} \right)^3 + \cdots + \frac{b}{n} \left( \frac{nb}{n} \right)^3 \\[9pt] &= \left( \frac{b}{n} \right)^4 (1^3 + 2^3 + \cdots + n^3). \end{align*}

  2. Proof. Next, we start with the given inequality

        \[ 1^3 + 2^3 + \cdots + (n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3, \]

    and multiply each term by \left( \frac{b}{n} \right)^4 to obtain

        \[ \left( \frac{b}{n} \right)^4 (1^3 + \cdots + (n-1)^3) < \frac{b^4}{4} < \left( \frac{b}{n} \right)^4 (1^3 + \cdots + n^3). \]

    Therefore, for all n \geq 1,

        \[ s_n < \frac{b^4}{4} < S_n. \]

    Now, to show that \frac{b^4}{4} is the only number that lies between s_n and S_n for all n \geq 1, we suppose A is any such number, i.e., s_n < A < S_n. We show that A < \frac{b^4}{4} and A > \frac{b^4}{4} both lead to contradictions, which means A = \frac{b^4}{4}.

    Suppose A < \frac{b^4}{4}, then

        \begin{align*} \frac{b^4}{4} < S_n && \implies && \frac{b^4}{4} - s_n &< S_n - s_n \\ && \implies && \frac{b^4}{4} - A &< \frac{b^4}{n} \\ && \implies && n &< \frac{b^4}{\frac{b^4}{4}- A}. \end{align*}

    But this cannot hold for all n \geq 1 since the positive integers are unbounded above.

    On the other hand, suppose A > \frac{b^4}{4}, then

        \begin{align*} A < S_n && \implies && A - s_n &< S_n - s_n \\ && \implies && A - \frac{b^4}{4} &< \frac{b^4}{n} \\ && \implies && n &< \frac{b^4}{A - \frac{b^4}{4}}. \end{align*}

    Again, this is a cannot hold for all positive integers n. Therefore, both A < \frac{b^4}{4} and A > \frac{b^4}{4} lead to contradictions, so we must have

        \[ A = \frac{b^4}{4}. \qquad \blacksquare \]

  3. Using parts (a) and (b) of this exercise and part (e) of the previous exercise we conclude that the area A under the curve ax^3 + c is given by

        \[ A = \frac{ab^4}{4} + bc. \]

2 comments

  1. Guilherme Coelho says:

    Thank you very much. Your solutions help a lot.

    Just a little detail: at part b., when analysing A < \frac{b^4}{4}, S_n - s_n equals \frac{b^4}{n}, not \frac{b^4}{4}.

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