Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

Consider the parabola \(y=x^{2}\) over the interval \([a, b]\), and let \(c=(a+b) / 2\) be the midpoint of \([a, b], d\) be the midpoint of \([a, c]\), and \(e\) be the midpoint of \([c, b]\). Let \(T_{1}\) be the triangle with vertices on the parabola at \(a, c\), and \(b\), and let \(T_{2}\) be the union of the two triangles with vertices on the parabola at \(a, d, c\) and \(c\), \(e, b\), respectively (Figure 14). Continue to build triangles on triangles in this manner, thus obtaining sets \(T_{3}, T_{4}, \ldots\). (a) Show that \(A\left(T_{1}\right)=(b-a)^{3} / 8\). (b) Show that \(A\left(T_{2}\right)=A\left(T_{1}\right) / 4\). (c) Let \(S\) be the parabolic segment cut off by the chord \(P Q\). Show that the area of \(S\) satisfies $$ A(S)=A\left(T_{1}\right)+A\left(T_{2}\right)+A\left(T_{3}\right)+\cdots=\frac{4}{3} A\left(T_{1}\right) $$ This is a famous result of Archimedes, which he obtained without coordinates. (d) Use these results to show that the area under \(y=x^{2}\) between \(a\) and \(b\) is \(b^{3} / 3-a^{3} / 3\).

Step-by-Step Solution

Answer

The area under the parabola between \(a\) and \(b\) is \(\frac{b^3}{3} - \frac{a^3}{3}\).

Step 1: Calculate the Area of Triangle T1

To find the area of triangle \( T_1 \), we use the vertices \( (a, a^2), (c, c^2), (b, b^2) \) where \( c = \frac{a+b}{2} \). The area of a triangle given vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is \( \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \). Substitute the points of \( T_1 \) to get the area: \[ A(T_1) = \frac{1}{2} \left| a(c^2 - b^2) + c(b^2 - a^2) + b(a^2 - c^2) \right|. \] With algebraic manipulation, this simplifies to \( A(T_1) = \frac{(b-a)^3}{8} \).

Step 2: Calculate the Area of Triangle T2 and Verify the Relation to T1

For \( T_2 \), we calculate the areas of two smaller triangles: the triangle with vertices \( (a, a^2), (d, d^2), (c, c^2) \) and the one with vertices \( (c, c^2), (e, e^2), (b, b^2) \). The midpoint \( d = \frac{a+c}{2} \), \( e = \frac{c+b}{2} \). Calculate each triangle's area similarly as in Step 1 and sum them to get \( A(T_2) \). After simplification, you'll find \( A(T_2) = \frac{1}{4}A(T_1)\).

Step 3: Show the Infinite Sum for the Parabolic Segment

The total area of the parabolic segment \( S \) can be represented as an infinite geometric series: \[ A(S) = A(T_1) + A(T_2) + A(T_3) + \cdots = \frac{4}{3} A(T_1). \]Each successive triangle area is \( \frac{1}{4} \) of the previous step, forming a geometric series \( A(T_1) \left(1 + \frac{1}{4} + \frac{1}{16} + \cdots\right) \). This converges to \( \frac{4}{3} A(T_1) \) because the series sum formula \( S = \frac{a_1}{1-r} \), where \( r = \frac{1}{4} \), applies.

Step 4: Derive the Area Under the Parabola Using Archimedes' Result

Using Archimedes' result, that the total area under the parabola between \( a \) and \( b \) (the region \( S \)) is \( \frac{4}{3} A(T_1) \). From Step 1, \( A(T_1) = \frac{(b-a)^3}{8} \). Substitute this into the formula for the area under the parabolic segment: \[ A(S) = \frac{4}{3} \left( \frac{(b-a)^3}{8} \right) = \frac{(b-a)^3}{6}. \]Thus, the area under the curve, \( y = x^2 \), is \( \frac{b^3}{3} - \frac{a^3}{3} \).

Key Concepts

Parabolic segment

A parabolic segment refers to the region in a plane that lies between a parabola and a line segment, called a chord. This concept is crucial to understanding how the area under a curve, particularly a parabolic curve like \( y = x^2 \), is calculated.
Archimedes was one of the first to explore this, using methods quite ahead of his time. In the context of the given exercise about finding these areas through triangles, the parabolic segment consists of areas of triangles that closely follow the curve as they are divided into smaller and smaller segments.

• The parabolic segment is important because it not only represents the actual region beneath the curve but exemplifies how geometric areas can approximate more complex shapes.
• The chord is the straight line each triangle shares with the parabola, while the parabolic arc is the curved boundary.

By calculating the areas of successive triangles, we can get very close to the true area of this segment, which is a concept that plays into integral calculus today.

Archimedes' method

Archimedes' method for determining the area of a parabolic segment is ingenious, involving geometric dissection. He essentially applied what we know today as an iterative process to compute areas.
To find the area of a parabolic segment, he constructed an infinite series of triangles underneath the parabola, starting with a single large triangle and then iteratively adding smaller triangles. Each triangle is constructed by taking midpoints, forming new triangles, and continuing this indefinitely.

• The significance of Archimedes’ method is in its ability to use simple shapes to approximate a much more complex figure.
• This method laid the groundwork for integral calculus by showing how infinite processes can be harnessed to calculate definite geometric areas.

Archimedes proved that the sum of these areas approaches a finite value, which can be calculated using a geometric series. Through this approach, he found that the area of a parabolic segment could be calculated as \( \frac{4}{3} \) times the area of the initial triangle.

Geometric series

A geometric series is a series with a constant ratio between successive terms. In the exercise, this concept is pivotal as we try to calculate the entire area of the parabolic segment.
The series arises from the observation that each successive set of triangles in Archimedes' method contributes a smaller area than the last. With each iteration, the newly added area becomes \( \frac{1}{4} \) of the previous area.

• The first term \( a_1 \) in this geometric series is the area of the first triangle \( T_1 \), and the common ratio \( r \) is \( \frac{1}{4} \).
• The series can be expressed as \( a_1 + a_1 \cdot \frac{1}{4} + a_1 \cdot \frac{1}{16} + \cdots \).

This series converges because the ratio \( r \) is less than 1. The sum \( S \) of this series is given by the formula \( S = \frac{a_1}{1-r} \). Applying this formula confirms that the area of the parabolic segment equals \( \frac{4}{3} A(T_1) \), showcasing the power of geometric series in solving problems of area under curves.