Problem 8 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

a. Compute: \(\mathrm{P}_{1} \quad \int_{2}^{4}\left[t \times t^{2}\right] d t\) and $$\mathrm{P}_{2} \int_{2}^{4} t d t \times \int_{2}^{4} t^{2} d t$$ b. Compute: \(\mathrm{Q}_{1}\) \(\int_{2}^{4} \frac{t^{3}}{t^{2}} d t \quad\) and \(\quad \mathrm{Q}_{2}\) $$\frac{\int_{2}^{4} t^{3} d t}{\int_{2}^{4} t^{2} d t}$$ c. What do these two problems illustrate?

Step-by-Step Solution

Answer

P₁ = 12, P₂ = 112; Q₁ = 6, Q₂ = 9/14. It illustrates integration properties and arithmetic operations with integrals.

Step 1: Simplify and Integrate P₁

First, simplify the integrand of \( P_1 \). We have \( \int_{2}^{4} [t \times t^2] \, dt = \int_{2}^{4} t^3 \, dt \). Using the power rule for integration, the antiderivative of \( t^3 \) is \( \frac{t^4}{4} \). The definite integral is then evaluated as:\[ \left[ \frac{t^4}{4} \right]_2^4 = \frac{4^4}{4} - \frac{2^4}{4} = 64/4 - 16/4 = 16 - 4 = 12 \].

Step 2: Integrate P₂ Components Separately

Now, compute \( P_2 = \left(\int_{2}^{4} t \, dt\right) \times \left(\int_{2}^{4} t^2 \, dt\right) \). Find each integral separately:- For \( \int_{2}^{4} t \, dt \), the antiderivative of \( t \) is \( \frac{t^2}{2} \). Evaluating the definite integral, \( \left[ \frac{t^2}{2} \right]_2^4 = \frac{4^2}{2} - \frac{2^2}{2} = 8 - 2 = 6 \).- For \( \int_{2}^{4} t^2 \, dt \), the antiderivative is \( \frac{t^3}{3} \). Evaluating, \( \left[ \frac{t^3}{3} \right]_2^4 = \frac{4^3}{3} - \frac{2^3}{3} = \frac{64}{3} - \frac{8}{3} = \frac{56}{3} \).Therefore, \( P_2 = 6 \times \frac{56}{3} = 112 \).

Step 3: Simplify and Integrate Q₁

For \( Q_1 \), simplify the integrand: \( \int_{2}^{4} \frac{t^3}{t^2} \, dt = \int_{2}^{4} t \, dt \). We have already computed that integral, which is \( 6 \).

Step 4: Evaluate Q₂ with Division of Integrals

For \( Q_2 \), compute \( \frac{\int_{2}^{4} t^3 \, dt}{\int_{2}^{4} t^2 \, dt} \). We have calculated these values:- \( \int_{2}^{4} t^3 \, dt = 12 \). - \( \int_{2}^{4} t^2 \, dt = \frac{56}{3} \).Thus, \( Q_2 = \frac{12}{\frac{56}{3}} = \frac{12 \times 3}{56} = \frac{36}{56} = \frac{9}{14} \).

Step 5: Determine the Illustration

Both problems involve integration but highlight different operational properties:- \( P_1 \) and \( Q_1 \) involve directly integrated simplified expressions.- \( P_2 \) and \( Q_2 \) demonstrate how combining integrals via multiplication or division can lead to different results.These examples illustrate the distributive property of integration versus the arithmetic manipulation of integrals.

Key Concepts

Definite Integral

A definite integral is an important concept in calculus, representing the accumulation of quantities such as areas under a curve. It is written in the form \( \int_{a}^{b} f(t) \, dt \), where \( a \) and \( b \) are the limits of integration. These limits define the interval over which you are summing the values of the function \( f(t) \).

When you evaluate a definite integral, you are essentially finding the "net area" between the function and the x-axis over the specified interval. This involves finding the antiderivative of the function, representing the continuous sum of values, and then applying the fundamental theorem of calculus to subtract the antiderivative at the lower limit from that at the upper limit.

Definite integrals have numerous practical applications, from calculating areas and volumes to solving differential equations. Understanding how to compute definite integrals is a foundational skill in calculus.

Power Rule for Integration

The power rule for integration is a straightforward method used to find the antiderivative of functions in the form of \( t^n \), where \( n eq -1 \). According to this rule: \[ \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \]where \( C \) is the constant of integration in indefinite integrals.

In the context of definite integrals, you can omit the constant since the subtraction of two antiderivatives at the bounds \( a \) and \( b \) cancels it out. This rule simplifies the computation of integrals considerably by providing a quick pathway to find antiderivatives for polynomial functions.

In our step-by-step solution, we applied the power rule when integrating terms like \( t^3 \) to find the total area under the curve by evaluating at the upper and lower bounds, thus simplifying and quickening the process.

Arithmetic Manipulation of Integrals

Arithmetic manipulation of integrals involves handling integrals through operations like addition, subtraction, multiplication, and division. It's important to understand these manipulations to effectively solve complex integral problems.

For example, in our exercise, multiple integrals were manipulated arithmetically in parts \( P_2 \) and \( Q_2 \). In \( P_2 \), we calculated each integral separately before multiplying their results, demonstrating multiplication of integrals. In \( Q_2 \), the results of two separate integrals were divided, underlining the operation of division on integrals.

This manipulation allows us to break down complex problems into manageable parts, calculate them independently, and then combine the results as needed. Careful attention to the rules of arithmetic ensures accurate results when manipulating integrals.

Distributive Property of Integration

The distributive property of integration is akin to the distributive law in basic arithmetic, allowing the distribution of integral operations over addition or subtraction within the integral’s scope. This property reveals the structure and relationships within complex expressions.

For instance, consider \( \int (f(t) + g(t)) \, dt \), which can be split into \( \int f(t) \, dt + \int g(t) \, dt \). This property is extremely useful when dealing with integrals of complex functions, simplifying them into smaller, more manageable integrals that are easier to compute.

In the exercise, this principle is highlighted by solving \( P_1 \) and \( Q_1 \) directly without combining different integral parts, unlike \( P_2 \) and \( Q_2 \) where different integrals are combined through multiplication and division. Understanding when and how to apply the distributive property is crucial in solving integration problems efficiently.