Problem 8 - Applied Calculus

Question

Find the integrals. $$\int(t+2) \sqrt{2+3 t} d t$$

Step-by-Step Solution

Answer

\( \frac{2}{45}(2+3t)^{5/2} + \frac{8}{27}(2+3t)^{3/2} + C \) is the integral of the given function.

Step 1: Identify the Substitution

Notice that the expression \( \int (t+2) \sqrt{2+3t} \, dt \) involves a composite function with an inner function \( 2 + 3t \) under the square root. We'll use substitution to simplify this integrand.

Step 2: Define the Substitution

Let \( u = 2 + 3t \). Consequently, differentiate this to get \( du = 3 \, dt \) or equivalently \( dt = \frac{du}{3} \).

Step 3: Change the Integrand

Substitute \( t \) and \( dt \) in terms of \( u \). Notice that \( t = \frac{u - 2}{3} \) from our substitution. Substitute back to the integral: \[ \int \left( \frac{u-2}{3} + 2 \right) \sqrt{u} \frac{du}{3} \].

Step 4: Simplify the Integral

Simplify the expression inside the integrand: \[ \int \left( \frac{u-2 + 6}{3} \right) \sqrt{u} \cdot \frac{du}{3} = \frac{1}{9} \int (u + 4) \sqrt{u} \, du \]. Now, split this into two integrals: \[ \frac{1}{9} \left( \int u \sqrt{u} \, du + 4 \int \sqrt{u} \, du \right) \].

Step 5: Solve the Individual Integrals

Compute each integral separately. For \( \int u \sqrt{u} \, du \): \[ \int u^{3/2} \, du = \frac{2}{5} u^{5/2} + C_1 \].For \( \int \sqrt{u} \, du \): \[ \int u^{1/2} \, du = \frac{2}{3} u^{3/2} + C_2 \].

Step 6: Combine the Results

Combine the results and place back in the original context:\[ \frac{1}{9} \left( \frac{2}{5} u^{5/2} + 4 \cdot \frac{2}{3} u^{3/2} \right) + C \].Simplify:\[ \frac{2}{45} u^{5/2} + \frac{8}{27} u^{3/2} + C \].

Step 7: Substitute Back for \( u \)

Finally, substitute \( u = 2 + 3t \) back into the solution:\[ \frac{2}{45} (2 + 3t)^{5/2} + \frac{8}{27} (2 + 3t)^{3/2} + C \]. This is the antiderivative of the original integrand.

Key Concepts

Integration by Substitution

Integration by substitution is an essential technique used in calculus to simplify the process of finding integrals, especially when dealing with complex or composite functions. The key idea is to transform a difficult integral into a simpler one that is easier to evaluate by substituting a part of the integrand with a new variable.

To begin with integration by substitution, follow these steps:

• Identify a part of the integrand that is suitable for substitution, typically an expression inside a function like a square root or another complicated function. In our example, we use the inner function \( 2 + 3t \).
• Define a substitution variable, \( u \), equal to this inner function (\( u = 2 + 3t \)).
• Calculate \( du \) by differentiating \( u \) with respect to \( t \), and express \( dt \) in terms of \( du \).
• Rewrite the whole integrand and the differential \( dt \) in terms of the new variable \( u \).
• Simplify and integrate with respect to \( u \).
• Don't forget to substitute back the original variable after integration.

By using substitution, the integral becomes much more manageable, as we swap the complex parts for easier expressions. This technique is a powerful tool in the integration toolbox, making it a favorite method for many problems in integral calculus.

Integral Calculus

Integral calculus is a fundamental part of calculus that focuses on the concept of integration, which is the reverse process of differentiation. Integration allows us to find the area under curves, the total accumulation of quantities, or to solve differential equations.

There are several types of integrals that students typically work with, including:

• Definite Integrals: Calculating the accumulation of quantities over an interval from \( a \) to \( b \), providing a numerical result.
• Indefinite Integrals: Finding the underlying function or family of functions whose derivative is the integrand. This is what we call the antiderivative.

Integral calculus is used in various fields such as physics, engineering, and economics to model real-world situations where calculating accumulated change is necessary. When performing integration, such as in our case \( \int (t+2)\sqrt{2+3t} \,dt \), we apply strategic techniques, including integration by substitution, to simplify complex expressions into solvable components. Mastery of these techniques is crucial for efficiently solving integration problems.

Antiderivative Calculations

In the realm of integral calculus, an antiderivative is a function whose derivative is the original function you have. When you are asked to find the integral of a function, you are essentially looking for its antiderivative.

To perform antiderivative calculations, follow these procesess:

• Identify the Function: Recognize the function that needs the antiderivative. In our example, it is the complex expression \( (t+2)\sqrt{2+3t} \).
• Apply Techniques: Use appropriate integration techniques such as substitution or parts to find the antiderivative. Breaking down the function into simpler components, like in previous steps, aids in identifying these techniques.
• Integrate Individual Components: Calculate the antiderivative for each part of your integrand separately if possible, then combine them as seen in steps 5 and 6 of the solution.
• Add Integration Constant: Since there could be multiple functions whose derivative matches the original function, always include an arbitrary constant, \( C \), in your final result.
• Back-Substitute: When using substitution methods, replace the substitute variable back into your final expression to revert to the original variable.

By executing these steps correctly, you form the solution to your integration problem, as seen with the antiderivative \[ \frac{2}{45} (2 + 3t)^{5/2} + \frac{8}{27} (2 + 3t)^{3/2} + C \]. Accurate antiderivative calculations are crucial for ensuring the correct interpretation of functions and their integrals.