Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \int_{1}^{2} \frac{d t}{\sqrt{t}+e} $$

Step-by-Step Solution

Answer

Evaluate the integral numerically as it simplifies to a more complex expression primarily involving logs and a constant.

Step 1: Set Up the Integral

This integral requires evaluation on the interval from 1 to 2 of the function \( \frac{1}{\sqrt{t} + e} \). Let's make sure we understand that our limits of integration are \( t = 1 \) and \( t = 2 \).

Step 2: Simplify the Denominator

Notice that the denominator \( \sqrt{t} + e \) is not easily separable directly. Instead, we can consider substituting \( u = \sqrt{t} \), which implies \( t = u^2 \) and \( dt = 2u \, du \). The limits of integration change accordingly from \( u=1 \) to \( u=\sqrt{2} \).

Step 3: Substitute Variables

By substituting \( u = \sqrt{t} \), the integral becomes: \( \int_{1}^{\sqrt{2}} \frac{2u}{u + e} \, du \). This substitution simplifies the denominator slightly and allows us to handle it more easily.

Step 4: Decompose the Fraction

Writing \( \frac{2u}{u+e} \) directly is slightly tricky, but let's rearrange to emphasize what treatment might work. Use long division or polynomial division techniques to simplify or reformat the fraction if possible. In this case, break it into \( 2 - \frac{2e}{u + e} \).

Step 5: Integrate Each Term Separately

Separate the integral into two different integrals: \( \int_{1}^{\sqrt{2}} 2 \, du - \int_{1}^{\sqrt{2}} \frac{2e}{u + e} \, du \). The first integral \( \int 2 \, du \) evaluates directly to \( 2u \), and the second integral can be expressed in terms of a logarithm.

Step 6: Evaluate the Integrals

Calculate each integral individually: - \( \int_{1}^{\sqrt{2}} 2 \, du = 2[u]_{1}^{\sqrt{2}} = 2(\sqrt{2} - 1) \).- \( \int_{1}^{\sqrt{2}} \frac{2e}{u + e} \, du = 2e [\ln |u+e|]_{1}^{\sqrt{2}} = 2e (\ln(\sqrt{2} + e) - \ln(1+e)) \).

Step 7: Combine Results

Combining these results gives us the final value:\[ 2(\sqrt{2} - 1) - 2e (\ln(\sqrt{2} + e) - \ln(1+e)) \] Evaluating this expression numerically provides the exact value of the definite integral.

Key Concepts

Integration Techniques

When approaching definite integrals, choosing the right technique can make a huge difference in solving them efficiently. Different methods can simplify complex expressions and make the entire process more manageable. Here are some popular techniques:

• Substitution: Often used when a function can be rewritten in terms of a simpler variable. It involves replacing one variable with another to transform the integral into an easier form.
• Integration by Parts: Useful for products of functions, this method is derived from the product rule for differentiation.
• Partial Fraction Decomposition: Used for rational functions, it breaks down complex fractions into simpler ones.

In our specific example, the substitution method was the best choice, as it managed to simplify the complicated expression involving a square root in the denominator.

Substitution Method

The substitution method in calculus is a powerful tool for tackling integrals that contain difficult expressions. The core idea is to change variables, replacing the challenging part of the function with a simpler expression. Let’s break it down:

• Select a substitution: Choose a new variable to represent a component of the integrand that simplifies the expression.
• Differential substitution: Substitute the differentials as well, ensuring the entire integral changes consistently.
• Change limits (if definite): Adjust the integration limits to reflect the substitution changes.

In the given exercise, the equation was \( rac{1}{ ext{denominator}} \) and the substitution \( u = \sqrt{t} \) helped reframe the integral, simplifying our calculations by moving the square root outside the complexity of the fraction.

Calculus

Calculus is a branch of mathematics that focuses on studying change through derivatives, integrals, and limits. It's essential for understanding how functions behave and solving a variety of mathematical problems.

• Derivatives: These measure the rate of change of a function. For instance, how a position changes with respect to time.
• Integrals: They are the inverse process to derivatives. While derivatives decompose change, integrals accumulate it, often calculating areas under curves or total quantities.
• Limits: These help understand the behavior of functions as inputs approach particular values, crucial for defining both derivatives and integrals.

The definite integral in our exercise calculates the net area or the cumulative sum of \( rac{1}{\sqrt{t} + e} \) from 1 to 2, showcasing the practical application of these concepts.

Mathematical Expressions

Mathematical expressions are combinations of symbols and variables that represent values and operations. They are the foundation of writing and solving equations.

• Variables: Symbols representing numbers or quantities, often denoted by letters like \( t \) or \( u \).
• Operations: Basic arithmetic operations (+, -, *, /) are frequently used in conjunction with higher-level functions like square roots, exponents, or logarithms.
• Complexity: Some expressions, like \( \sqrt{t} + e \), involve multiple operations and constants that require specific techniques to simplify or solve.

In the example provided, rewriting the denominator \( \sqrt{t} + e \) using the substitution variable \( u \) transformed a complex expression into something more approachable, highlighting the importance of mastering these theoretical tools.