Problem 44 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(25-46,\) use substitution to evaluate the integral. $$\int \frac{d x}{\sqrt{5 x+8}}$$

Step-by-Step Solution

Answer

The integral of \( \frac{1}{\sqrt{5x+8}} \) dx is \( \frac{2}{5} \sqrt{5x+8} + C \)

Step 1: Choose a new variable u

Choosing a new variable \(u\) can make the problem easier to solve. Here, let's say \(u = 5x + 8\). This makes the denominator inside the square root become \(u\). Remember to substitute \(u\) back into the integral in later steps.

Step 2: Find du by taking derivative of u

Taking the derivative of \(u = 5x + 8\) with respect to \(x\), yields \(du = 5 dx\). We isolate \(dx\) for substitution in the integral and obtain \(dx = du/5\).

Step 3: Substitute u and du in integral

Substituting \(u\) and \(dx\) into the original integral, we get: \(\int \frac{du/5}{\sqrt{u}}=\frac{1}{5}\int u^{-1/2} du\).

Step 4: Evaluate the integral

Now we can evaluate the integral using the antiderivative of \(u^{-1/2}\), which is \(2u^{1/2}\). Doing so, we find the integral equals \(\frac{1}{5}(2u^{1/2}) = 2/5 \sqrt{u}\).

Step 5: Substitute u back

Substitute \(u\) back into the equation, which we originally stated as \(u = 5x + 8\). Our final result is : \(2/5 \sqrt{5x+8}\), plus a constant, say \(C\). So, the result is \(2/5 \sqrt{5x+8}+C\).

Key Concepts

Substitution Method

The substitution method is a powerful technique in calculus, especially for integrals involving complex expressions. This approach simplifies integrals by converting them into a more manageable form. In the provided example, we replace a bulky expression under the square root with a simple variable, say \( u \). Here's how it's done:

• Identify a part of the integrand that, when substituted, simplifies the integral. In this case, \( u = 5x + 8 \) is chosen.
• Express \( dx \) in terms of \( du \), which involves differentiating the substitution expression \( u \) with respect to \( x \). Thus, we get \( dx = \frac{du}{5} \).
• Substitute these into the integral, effectively transforming the original variable \( x \) based integral into one in terms of \( u \).

Using substitution allows more straightforward integration, turning difficult forms into basic ones you can easily integrate.

Definite Integral

While the original problem focuses on an indefinite integral, let's explore definite integrals. They compute the total accumulation of quantity or the area under a curve between two limits. Unlike the indefinite integral, which results in a general form plus a constant \( C \), the definite integral provides a numerical answer.

• To solve a definite integral using substitution, first redefine the variable as explained earlier.
• Then, adjust the limits of integration according to the substitution. For example, convert \( x \)-limits into \( u \)-limits by substituting them into \( u = 5x + 8 \).
• Finally, evaluate the integral with these new limits to find exact values.

This method ensures precision, especially useful in practical applications like physics and engineering, where specific solutions are often needed.

Antiderivative

Finding the antiderivative is the core of solving integrals. The antiderivative, often known as the indefinite integral, is a function whose derivative is the original function in the integrand. For instance, to solve \( \int u^{-1/2} \, du \), we identify the antiderivative as \( 2u^{1/2} \).Here's why it's vital:

• It provides the general solution to differential equations and integrals.
• Understanding antiderivatives helps you find functions from their rates of change.
• They are fundamental in reversing differentiation, thus helping solve area problems.

Working out antiderivatives requires knowing basic derivative rules and is crucial for integration, simplifying many mathematical problems.

Integral Calculus

Integral calculus studies integrals and their properties. It's one of the main branches of calculus alongside differential calculus. While differential calculus focuses on rates of change, integral calculus is about accumulation - think of it like gathering everything up. Some key purposes and uses include:

• Calculating areas under curves, which represent quantities like distance and probability.
• Finding accumulated values over intervals, such as total growth or economic quantities.
• Solving complex problems in physics, engineering, and probability by providing precise accumulation functions.

Integral calculus uses several methods, like substitution (used in the exercise), integration by parts, or partial fractions to tackle a wide range of integrals.