Problem 39 - Calculus

Question

Use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{5}^{8} \sqrt{3 x+1} d x $$

Step-by-Step Solution

Answer

The definite integral evaluates to \( \frac{122}{9} \).

Step 1: Identify the Function for Substitution

To use substitution, identify a part of the integrand that can be substituted with a single variable. In the integral \( \int_{5}^{8} \sqrt{3x+1} \, dx \), the expression \( u = 3x + 1 \) is a good choice for substitution as it simplifies the square root.

Step 2: Differentiate to Find du

Differentiate the substitution to find \( du \). Since \( u = 3x + 1 \), differentiating gives \( \frac{du}{dx} = 3 \) or \( du = 3 \, dx \). We can write \( dx = \frac{du}{3} \).

Step 3: Change Limits of Integration

Substitute the limits of \( x \) into \( u = 3x + 1 \). For \( x = 5 \), \( u = 3(5) + 1 = 16 \). For \( x = 8 \), \( u = 3(8) + 1 = 25 \). The limits are now from \( u = 16 \) to \( u = 25 \).

Step 4: Substitute and Simplify the Integral

Substitute \( u \) and \( du \) into the integral: \[ \int_{5}^{8} \sqrt{3x+1} \, dx = \int_{16}^{25} \sqrt{u} \cdot \frac{1}{3} \, du = \frac{1}{3} \int_{16}^{25} u^{1/2} \, du. \]

Step 5: Evaluate the Integral

Now integrate \( \frac{1}{3} \int u^{1/2} \, du \) as follows:\[ \frac{1}{3} \left[ \frac{u^{3/2}}{3/2} \right]_{16}^{25} = \frac{1}{3} \cdot \frac{2}{3} \left[ u^{3/2} \right]_{16}^{25}. \] This simplifies to \( \frac{2}{9} \left[ 25^{3/2} - 16^{3/2} \right]. \)

Step 6: Calculate the Final Answer

Calculate the values of \( 25^{3/2} \) and \( 16^{3/2} \). \( 25^{3/2} = (25^{1/2})^3 = 5^3 = 125 \). \( 16^{3/2} = (16^{1/2})^3 = 4^3 = 64 \). So, the final calculation becomes:\[ \frac{2}{9} (125 - 64) = \frac{2}{9} \times 61 = \frac{122}{9}. \] This simplifies to approximately 13.56 or exactly as \( \frac{122}{9} \).

Key Concepts

Integration Techniques

Integration techniques are a set of methods used to solve integrals, which are key elements of calculus. Among these techniques, substitution is a very popular one.

• Substitution is ideal when the integral contains a composite function, where one function is nested within another.
• The goal is to simplify the integral by replacing a part of the function with a new variable, resulting in a simpler integral to solve.

Remember: mastering integration techniques is crucial because they help us solve a wide variety of practical and theoretical problems in mathematical analysis.

Definite Integrals

Definite integrals represent the accumulation of quantities, like areas under curves, between two specified points. They differ from indefinite integrals, which find general antiderivatives without specific limits.

• A definite integral is expressed as \( \int_{a}^{b} f(x) \, dx \), with limits of integration from \( a \) to \( b \).
• They provide a numerical value which can represent quantities like total distance, area, or volume, depending on the context of the problem.

Understanding definite integrals is essential because they allow us to calculate these accumulative quantities with respect to continuous change.

U-Substitution

U-substitution is a powerful technique akin to reverse chain rule for integration. It's often used to simplify integrals by reassigning a part of the function to a new variable, \( u \). Step by step:- Identify a subexpression within the integral that can be simplified. In our case, it's \( u = 3x + 1 \).- Differentiate \( u \) to find \( du \), making it easier to express the original integral in terms of \( u \).Since \( du = 3 \, dx \), we can express \( dx \) as \( \frac{du}{3} \) and reformulate the integral. U-substitution transforms complex integrals into simpler forms, making them easier and quicker to solve.

Limits of Integration

When substitution is applied to definite integrals, the limits of integration must be adjusted to match the new variable. This step ensures that the boundaries of the integration remain consistent.

• Initial limits \( x = 5 \) and \( x = 8 \) are transformed according to the substitution formula \( u = 3x + 1 \).
• Thus, \( x = 5 \) becomes \( u = 16 \) and \( x = 8 \) becomes \( u = 25 \).

These transformed limits are used in the new expression of the integral, ensuring the final calculated value reflects the correct interval of accumulation. Correct limit transformations are crucial for accurate results in definite integral calculations.