Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 35-62, use the Substitution Rule for Definite Integrals to evaluate each definite integral. \(\int_{-3}^{3} \sqrt{7+2 t^{2}}(8 t) d t\)

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Identify the Substitution

We want to integrate \( \int_{-3}^{3} \sqrt{7+2t^2}(8t) \, dt \). We can use a substitution method here by letting \( u = 7 + 2t^2 \). Hence, the derivative \( du = 4t \, dt \).

Step 2: Solve for dt

From the substitution \( u = 7 + 2t^2 \), we have \( du = 4t \, dt \), which implies \( dt = \frac{du}{4t} \). Our integral contains an \( 8t \, dt \) term, which can be rewritten using this substitution.

Step 3: Substitute u and Adjust the Integral

The integral \( 8t \, dt \) can be rewritten in terms of \( du \) as \( 2 \, du \) using \( dt = \frac{du}{4t} \). Hence, the integral becomes \( \int_{?}^{?} \sqrt{u} \cdot 2 \, du \).

Step 4: Change the Limits of Integration

When \( t = -3 \), \( u = 7 + 2(-3)^2 = 25 \). When \( t = 3 \), \( u = 7 + 2(3)^2 = 25 \). Therefore, the limits of integration are from 25 to 25.

Step 5: Evaluate the New Integral

Evaluate \( \int_{25}^{25} \sqrt{u} \cdot 2 \, du \), which equals 0 because the limits of integration are the same.

Key Concepts

Integral Calculus

Integral calculus is all about finding the total or accumulated quantity when a function varies over a range. Unlike derivatives, which focus on rates of change, integrals focus on accumulation. This branch of calculus helps us calculate areas under curves, volumes of solids of revolution, and many other quantities that can be expressed in terms of a summation. There are two main types of integrals: definite and indefinite. Indefinite integrals yield a function (the antiderivative) without boundaries, while definite integrals provide the accumulated value over a specific interval. The definite integral \[ \int_{a}^{b} f(x) \, dx \] represents the area between the curve \( f(x) \), the x-axis, and the vertical lines \( x=a \) and \( x=b \). In our exercise, integral calculus is used to simplify and solve a definite integral by applying the substitution rule.

Definite Integrals

Definite integrals are a key concept in integral calculus, often representing the area under a curve from one point to another. Mathematically, a definite integral is defined as the limit of a sum of areas of rectangles as the width of the rectangles approaches zero. - They are express as: \[ \int_{a}^{b} f(x) \, dx \]- The "a" and "b" are called the limits of integration and determine the interval over which the area is calculated. - A property of definite integrals is that if the upper and lower limits are the same (i.e., \( a = b \)), the integral evaluates to zero, because there is no interval.
In the provided exercise, due to the substitution, both limits became 25, resulting in an integral of zero, which intuitively makes sense because no area is enclosed when both bounds are the same.

U-Substitution

U-substitution is a technique used in calculus to ease the process of integrating functions. It's particularly useful when dealing with definite and indefinite integrals that involve composite functions, often allowing us to transform an unmanageable integral into a simpler one. Here’s the basic process:

• Identify a part of the integrand that can be substituted. Usually, this is the inner function in a composite function, such as \( u = g(x) \).
• Calculate the derivative \( du \) and express \( dx \) in terms of \( du \).
• Substitute \( u \) and \( du \) in the integral, simplifying if needed.
• Adjust the limits of integration; these must match the new variable \( u \).

In the exercise, by letting \( u = 7 + 2t^2 \), we streamlined the integration by transforming \( 8t \, dt \) to \( 2 \, du \). This simpler form is often much easier to evaluate. Always remember to change back to the original variable if you're working with indefinite integrals or if the bounds of integration weren't changed properly.

Limits of Integration

The limits of integration in a definite integral, noted as \( a \) and \( b \) in the integral notation \( \int_{a}^{b} f(x) \, dx \), dictate the interval over which you are calculating the accumulation of the function \( f(x) \). These values are crucial as they define the "start" and "end" points of the area or quantity being calculated.When performing u-substitution in definite integrals, it's important to convert these limits to the new variable \( u \).

• Substitute the original limits into the \( u \) function to find the new bounds.
• Re-evaluate the transformed integral using these new limits.

In our exercise, substituting for \( u \) resulted in the same start and end point, 25, which means the integral is calculated over a zero-width interval, directly giving a result of zero irrespective of the function. Understanding how to change limits of integration when using u-substitution is essential for correct integral evaluation.