Question

Consider the curve \(y=1 / x^{2}\) for \(1 \leq x \leq 6\). (a) Calculate the area under this curve. (b) Determine \(c\) so that the line \(x=c\) bisects the area of part (a). (c) Determine \(d\) so that the line \(y=d\) bisects the area of nart (a).

Short answer

Area is \(\frac{5}{6}\); \(c = \frac{12}{7}\); \(d = \frac{49}{144}\).

Step-by-step solution

Define the Integral for Area

To find the area under the curve \( y = \frac{1}{x^2} \) from \( x=1 \) to \( x=6 \), set up the definite integral: \[ A = \int_{1}^{6} \frac{1}{x^2} \, dx \].

Compute the Integral

Integrate \( \frac{1}{x^2} \). Since \( \frac{1}{x^2} = x^{-2} \), the integral becomes \( \int x^{-2} \, dx = -x^{-1} + C \). Evaluate from 1 to 6: \[ A = \left[ -\frac{1}{x} \right]_{1}^{6} = -\frac{1}{6} - (-1) = \frac{5}{6} \].

Solve for c to Bisect Area

We want \( \int_{1}^{c} \frac{1}{x^2} \, dx = \int_{c}^{6} \frac{1}{x^2} \, dx = \frac{5}{12} \). Compute the integral: \[ \int_{1}^{c} \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_{1}^{c} = -\frac{1}{c} + 1 \]. Set \(-\frac{1}{c} + 1 = \frac{5}{12}\) to find \(c\). Solving gives \[ c = \frac{12}{7} \].

Solve for d to Bisect Area in y-direction

We need to find \( d \) such that the area from \( y = \frac{1}{x^2} \) to \( y = d \) equals \( \frac{5}{12} \). This requires integrating horizontally, with \( x = \sqrt{\frac{1}{y}} \). So, \( A_y = \int_{d}^{1} \frac{1}{\sqrt{y}} \, dy \) should equal \( \frac{5}{12} \). Compute the integral: \[ A_y = \left[ -2y^{1/2} \right]_{d}^{1} = -2 + 2\sqrt{d} \]. Set \(-2 + 2\sqrt{d} = \frac{5}{12}\) and solve for \( d \), yielding \[ d = \frac{49}{144} \].

Key concepts

Area under a curve

The concept of finding the "area under a curve" is a fundamental aspect of calculus. When you want to find the total area beneath a curve, say from one point to another on the x-axis, you actually need to compute the integral of the function that represents the curve. For the curve given by the function \( y = \frac{1}{x^2} \), we compute this as a definite integral.

A definite integral is written as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration, and \( f(x) \) is your function. In this case, you want the integral from \( x = 1 \) to \( x = 6 \).

This process involves:

• Setting up the integral: \( \int_{1}^{6} \frac{1}{x^2} \, dx \).
• Finding the antiderivative of \( \frac{1}{x^2} \), which is \(-\frac{1}{x} + C\).
• Evaluating this antiderivative from 1 to 6, resulting in \( \frac{5}{6} \), which is the area under the curve.

Definite integral

A 'definite integral' is a key concept in calculus. This type of integral computes the accumulated value of a function over a specified interval. Unlike indefinite integrals, definite integrals result in a specific numerical value. This value represents the signed area between the curve and the x-axis between two given points on the x-axis.

To calculate the definite integral of a function \( f(x) \) over an interval \([a, b]\), follow these steps:

• Find the antiderivative \( F(x) \) of the function \( f(x) \).
• Substitute the upper limit \( b \) into \( F(x) \), and subtract the result when you substitute the lower limit \( a \) into \( F(x) \).
• This calculation is often represented as: \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).

For the curve \( y = \frac{1}{x^2} \), calculating the definite integral from \( x = 1 \) to \( x = 6 \) involves finding \( \left[-\frac{1}{x}\right]_{1}^{6} \), which gives us an area of \( \frac{5}{6} \).

This technique is powerful because it allows us to determine the exact area contained under a curve within a specified range.

Bisecting area

"Bisecting area" refers to the process of cutting the region under a curve into two equal parts, either horizontally or vertically. In calculus problems, this is often done to find a specific point on the x-axis or a specific value on the y-axis where this bisecting line should be drawn.

When bisecting vertically, we want to determine \( c \) such that the area from \( x = 1 \) to \( x = c \) is equal to the area from \( x = c \) to \( x = 6 \). You find this by setting the integral \( \int_{1}^{c} \frac{1}{x^2} \, dx \) equal to half of the total area, \( \frac{5}{12} \). Solving \(-\frac{1}{c} + 1 = \frac{5}{12}\) gives \( c = \frac{12}{7} \).

For horizontal bisection, we locate \( d \) such that the area between \( y = \frac{1}{x^2} \) and \( y = d \) equals the other half of the area under the curve. This involves integrating with respect to \( y \) and solving \( \int_{d}^{1} \frac{1}{\sqrt{y}} \, dy = \frac{5}{12} \). Solving results in \( d = \frac{49}{144} \).

These methods of bisecting areas are useful for understanding symmetry and balance in areas under curves, and have practical uses in fields that require equal distribution.