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Calculate the area under higher power curves

by RoRi

Given the inequality

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

(valid for all integers n, k \geq 1), generalize the results of Exercise I.4.2 to calculate x^n for any integer n.

Using the same method as in the previous two exercise, here and here, we first calculate the lower and upper sums by adding up the areas of the rectangles below and above the curve, respectively.

    \begin{align*} s_n &= \frac{b}{n} \left( \frac{b}{n} \right)^k + \frac{b}{n} \left( \frac{2b}{n} \right)^k + \cdots + \frac{b}{n} \left( \frac{(n-1)b}{n} \right)^k \\[9pt] &= \left( \frac{b}{n} \right)^{k+1} (1^k + 2^k + \cdots + (n-1)^k),\\ \intertext{and,} S_n &= \frac{b}{n} \left( \frac{b}{n} \right)^k + \frac{b}{n} \left( \frac{2b}{n} \right)^k + \cdots + \frac{b}{n} \left( \frac{nb}{n} \right)^k \\[9pt] &= \left( \frac{b}{n} \right)^{k+1} (1^k + 2^k + \cdots + n^k). \end{align*}

Then, starting with the given inequality

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

we multiply each term by \left( \frac{b}{n} \right)^{k+1} \right) to obtain

    \[ s_n < \frac{b^{k+1}}{k+1} < S_n. \]

Following the same method as in the previous exercise we then obtain that if the ordinate at each x is given by ax^k + c the area A is given by

    \[ A = \frac{ab^{k+1}}{k+1} + bc. \]

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

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

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

by RoRi

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

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