Problem 11 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Write an integral that is the area of the region bounded by the graphs of a. \(y=t^{2}-t \quad\) and \(\quad y=0, \quad t=1, \quad t=2\). b. \(y=t^{2}, \quad\) and \(\quad y=t, \quad t=1, \quad t=2\). c. \(y=t^{2}, \quad\) and \(\quad y=t, \quad t=0, \quad t=1\). d. \(y=2 \times t^{5}-t^{4},\) and \(\quad y=0, \quad t=1, \quad t=2\). e. \(y=2 \times t^{5}, \quad\) and \(\quad y=t^{4}, \quad t=1, \quad t=2\). It is useful to sketch the regions.

Step-by-Step Solution

Answer

The integrals for a-e are: (a) \(\int_{1}^{2} (t^2-t) \, dt \), (b) \(\int_{1}^{2} (t-t^2) \, dt \), (c) \(\int_{0}^{1} (t-t^2) \, dt \), (d) \(\int_{1}^{2} (2t^5-t^4) \, dt \), (e) \(\int_{1}^{2} (2t^5-t^4) \, dt \).

Step 1: Identify the Curves for Part (a)

The functions given are \( y = t^2 - t \) and \( y = 0 \). The region is bounded by these curves and the vertical lines \( t = 1 \) and \( t = 2 \). The region lies above the x-axis and below the parabola, between \( t = 1 \) and \( t = 2 \).

Step 2: Set Up the Integral for Part (a)

The area under the curve from \( t = 1 \) to \( t = 2 \) above \( y = 0 \) is represented by the integral: \[ \int_{1}^{2} (t^2 - t) \, dt \].

Step 3: Identify the Curves for Part (b)

The functions are \( y = t^2 \) and \( y = t \). The bounded region is where \( y = t^2 \) is below \( y = t \) between \( t = 1 \) and \( t = 2 \).

Step 4: Set Up the Integral for Part (b)

The area of the region is given by the integral \[ \int_{1}^{2} (t - t^2) \, dt \], where \( y = t \) is subtracted from \( y = t^2 \).

Step 5: Identify the Curves for Part (c)

The curves are \( y = t^2 \) and \( y = t \) with bounds at \( t = 0 \) and \( t = 1 \). Here, \( y = t^2 \) is above \( y = t \) in this interval.

Step 6: Set Up the Integral for Part (c)

The bounded area is \[ \int_{0}^{1} (t - t^2) \, dt \], as \( y = t \) is subtracted from \( y = t^2 \).

Step 7: Identify the Curves for Part (d)

The functions given are \( y = 2t^5 - t^4 \) and \( y = 0 \). The area in question is above the x-axis from \( t = 1 \) to \( t = 2 \).

Step 8: Set Up the Integral for Part (d)

The area of the region is \[ \int_{1}^{2} (2t^5 - t^4) \, dt \].

Step 9: Identify the Curves for Part (e)

The curves are \( y = 2t^5 \) and \( y = t^4 \), with \( t = 1 \) and \( t = 2 \) as bounds. The curve \( y = 2t^5 \) lies above \( y = t^4 \) in this interval.

Step 10: Set Up the Integral for Part (e)

The area is given by \[ \int_{1}^{2} (2t^5 - t^4) \, dt \], representing the region between the curves.

Key Concepts

Area Under a Curve

The concept of finding the area under a curve is a fundamental application of definite integrals. When you look at a graph, the curve represents a function, usually denoted as \( y = f(t) \). The area under this curve from one point to another is the space between the curve and the x-axis, bounded by vertical lines at two limits, say \( t = a \) and \( t = b \).
To find this area, we use the integral symbol, which adds up an infinite number of infinitesimally small rectangles under the curve. The resulting integral looks like this: \[\int_{a}^{b} f(t) \, dt.\]
This notion allows us to calculate various quantities, such as the distance traveled over time or, in engineering, the work done by a force. It's a versatile concept in mathematics that applies to countless scenarios where a continuous change is involved.
The challenge often lies in correctly identifying the function \( f(t) \) you need to integrate and setting the proper bounds \( a \) and \( b \) for your problem.

Bounded Regions

In calculus, bounded regions refer to the enclosed area between two or more curves over a given interval. This is often visualized by sketching the curves to see where they intersect and mark the segments where one curve is above the other. For example, given two functions, \( y = t^2 \) and \( y = t \), to find the area of the bounded region between those curves, we need to
1. Determine the points of intersection to set the integral limits, e.g., \( t = 1 \) to \( t = 2 \). 2. Identify which function forms the upper boundary and which the lower within this interval, e.g., \( y = t \) might be above \( y = t^2 \). 3. Set up the definite integral by subtracting the lower function from the upper function, like this:
\[\int_{1}^{2} (t - t^2) \, dt.\]
This subtraction finds the area between the two curves by effectively "slicing" the space into vertical sections and summing these volumes.

Integration

Integration is the mathematical process used to find areas, volumes, central points, and many useful things. It is the reverse process of differentiation. In practice, integration sums all small changes that occur in a function over an interval, providing a cumulative total amount.
The integral of a function \( f(t) \) either over its entire domain or between specified limits \( t = a \) and \( t = b \), is denoted as \( \int f(t) \, dt \). This symbol signifies the process of integrating.
There are two main types of integrals: indefinite and definite integrals:

• Indefinite Integrals: Represent a family of functions and include an arbitrary constant \( C \). For example, the integral of \( t \) is \( \frac{1}{2}t^2 + C \).
• Definite Integrals: Calculate the net area under a curve between two points. These are used to evaluate the actual area of bound regions, as discussed. Such as \( \int_{1}^{2} t^2 \, dt \).

Integration can be seen as a powerful toolkit to solve problems involving total accumulation of change and is crucial for analyzing systems that evolve over time.