Problem 58 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int_{1}^{4} \frac{8 \cosh \sqrt{x}}{\sqrt{x}} d x$$

Step-by-Step Solution

Answer

The value of the integral is \( 16(\sinh(2) - \sinh(1)) \).

Step 1: Identify the substitution

The integral contains the term \( \cosh \sqrt{x} \) and \( \sqrt{x} \), suggesting a substitution. Let \( u = \sqrt{x} \), hence \( x = u^2 \).

Step 2: Differentiate the substitution

Since \( u = \sqrt{x} \), the differential \( du = \frac{1}{2\sqrt{x}} dx \). By rearranging we find \( dx = 2u \, du \).

Step 3: Change the limits of integration

With the substitution \( u = \sqrt{x} \): when \( x = 1 \), \( u = \sqrt{1} = 1 \); when \( x = 4 \), \( u = \sqrt{4} = 2 \). So, the limits change from \( x = 1 \) to 4 to \( u = 1 \) to 2.

Step 4: Substitute and simplify the integral

Substitute \( u \) and \( dx = 2u \, du \) into the integral:\[\int_{1}^{2} \frac{8\cosh(u)}{u} \cdot 2u \, du = \int_{1}^{2} 16\cosh(u) \, du.\]The \( u \) terms cancel out.

Step 5: Integrate the function

The integral is now \( \int 16 \cosh(u) \, du \). The integral of \( \cosh(u) \) is \( \sinh(u) \). Thus, evaluate:\[16\sinh(u) \Big|_{1}^{2} = 16\left( \sinh(2) - \sinh(1) \right).\]

Step 6: Evaluate the definite integral

Calculate \( 16 \left( \sinh(2) - \sinh(1) \right) \) to obtain the final value. Use a calculator if necessary to find the exact numerical values for \( \sinh(2) \) and \( \sinh(1) \).

Key Concepts

Substitution Method

The substitution method is a powerful technique used to simplify complex integrals, making them easier to evaluate. In this method, a new variable, often denoted as \( u \), is introduced to replace a more complicated part of the integrand. This transformation can make integration straightforward.To use substitution:

• Identify a part of the integral that can be replaced with a single variable \( u \), so differentiation of \( u \) makes the integral manageable.
• Find the derivative of \( u \) with respect to \( x \) to get \( du \). Substitute \( dx \) in terms of \( du \).
• Change the limits of integration to reflect the new variable.
• Transform the integral to the \( u \) variable and solve it.

In our given exercise, substituting \( u = \sqrt{x} \) simplifies the integral. By changing variables, we also simplified the process of evaluating the definite integral.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola, much like how sine and cosine relate to a circle. The most common hyperbolic functions are \( \sinh \), \( \cosh \), and \( \tanh \).Key properties of hyperbolic functions:

• The hyperbolic cosine function, \( \cosh(x) \), is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).
• The hyperbolic sine function, \( \sinh(x) \), is \( \sinh(x) = \frac{e^x - e^{-x}}{2} \).
• These functions are helpful when dealing with integrals that involve expressions like \( \cosh(\sqrt{x}) \).

In our context, integrating \( \cosh(u) \) results in \( \sinh(u) \), which mirrors how integrating \( \cos(x) \) yields \( \sin(x) \). Understanding these functions is crucial when an integral contains such terms.

Limits of Integration

Limits of integration define the interval over which the integral is evaluated for definite integrals. When using substitution, these limits must be adjusted to match the new variable introduced.Steps to change the limits:

• Initially, identify the original limits for the variable \( x \).
• Using the substitution, calculate the new limits in terms of \( u \). For example, if \( x = a \) and \( x = b \) are the original limits, find \( u \) by substituting into the equation \( u = \sqrt{x} \).
• Apply these new limits of integration to solve the integral in terms of the newly defined variable.

In this exercise, when \( x \) is from 1 to 4, substituting \( u = \sqrt{x} \) changes the limits to \( u \) from 1 to 2. Correctly adjusting these limits is essential for calculating a definite integral accurately.