Problem 42 - Calculus

Question

Use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{1}^{3} \frac{x^{2}+1}{\sqrt{x^{3}+3 x}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{8}{3} \).

Step 1: Identify the Inner Function

First, identify the part of the integrand that affects the expression under the square root. Let's choose the expression inside the square root as our inner function: \( u = x^3 + 3x \).

Step 2: Differentiate the Inner Function

Find the derivative of \( u \) with respect to \( x \). Calculate \( \frac{du}{dx} = 3x^2 + 3 \), hence \( du = (3x^2 + 3) dx \).

Step 3: Rewrite the Integral

To use substitution, we need to match \( du \) in the original integral. Observe: \( \frac{x^2 + 1}{\sqrt{x^3 + 3x}} \) then simplifies using \( du = (3x^2 + 3)dx = 3(x^2+1)dx \), therefore, \( dx = \frac{1}{3} \frac{du}{x^2+1} \).

Step 4: Change Limits of Integration

Substitute the original limits in terms of \( u \). When \( x = 1 \), \( u = 1^3 + 3 \times 1 = 4 \). When \( x = 3 \), \( u = 3^3 + 3 \times 3 = 36 \). The integral in terms of \( u \) becomes: \( \int_{4}^{36} \frac{1}{3\sqrt{u}} du \).

Step 5: Evaluate the Integral

Factor out the constant \( \frac{1}{3} \), giving \( \frac{1}{3} \int_{4}^{36} u^{-\frac{1}{2}} du \). Integrate \( u^{-\frac{1}{2}} \) to get \( 2u^{\frac{1}{2}} \), so the result becomes: \( 2 \cdot \frac{1}{3} \cdot [u^{\frac{1}{2}}]_{4}^{36} = \frac{2}{3}([36^{\frac{1}{2}}] - [4^{\frac{1}{2}}]) \).

Step 6: Substitute the Limits and Simplify

Calculate \( 36^{\frac{1}{2}} = 6 \) and \( 4^{\frac{1}{2}} = 2 \). The expression simplifies to \( \frac{2}{3}(6-2) = \frac{2}{3} \times 4 = \frac{8}{3} \).

Key Concepts

Definite Integrals

Definite integrals are a way to find the total accumulation of a quantity over an interval. They differ from indefinite integrals, which do not have set limits and represent a family of functions. In the context of a definite integral, we define two limits of integration that specify the range over which we are summing the area under a curve.

• The lower limit, represented as the bottom number in the integral sign, marks the starting point of the interval.
• The upper limit, shown as the top number, indicates the endpoint of the interval.

Think of definite integrals as a tool to find areas under curves between two points on a graph. They also have practical applications in physics and engineering, such as calculating distances from speeds or determining quantities from rates.

Inner Function

The inner function is a crucial component when using the substitution rule for integrals. It is the part of the integrand (the function being integrated) that is transformed into a new variable to simplify the integration process. Here's how it works:

• Identify the part of the integrand that fits under another function or operation, like a square root.
• Assign this part a new variable, often denoted as \( u \). This turns complex expressions into simpler forms.

In our original exercise, the inner function is \( u = x^3 + 3x \). By transforming this expression, the integration becomes more manageable. This approach is akin to "chunking" complex problems into digestible parts to make them easier to solve.

Limits of Integration

When performing definite integrals, switching between the original and new variables requires adjusting the limits of integration. After identifying and differentiating the inner function, convert the original variable's limits into those for the new variable, \( u \). Follow these steps:

• Substitute each limit, from the original variable expression into the inner function equation.
• Calculate the corresponding values for these limits using the defined inner function.

In the given problem, for \( x = 1 \), \( u = 4 \) and for \( x = 3 \), \( u = 36 \). These new limits transform the integral into a simpler form. They are like translating coordinates from one map to another, ensuring you're accurately representing the integration region.

Integrand

The integrand is the function you are integrating in an integral problem. It plays a vital role in determining the approach for solving an integral, particularly when using substitution or other advanced techniques. Here’s why it matters:

• Assessing the structure of the integrand helps in choosing the best method to simplify and evaluate the integral.
• In substitution, parts of the integrand are selectively transformed to a more convenient form, utilizing the inner function.

For our exercise, the integrand is \( \frac{x^2+1}{\sqrt{x^3+3x}} \). By breaking it down into simpler parts, specifically with the substitution \( u = x^3 + 3x \), integrating becomes straightforward. Visualize the integrand as the landscape you navigate during an integration journey, where simplification practices clear a path for easier travel.