Question

In Problems 1-16, perform the indicated integrations. \(\int t(3 t+2)^{3 / 2} d t\)

Short answer

\( \frac{2}{63} (3t + 2)^{7/2} - \frac{4}{45} (3t + 2)^{5/2} + C \)

Step-by-step solution

Choose a suitable substitution

To simplify the integration, we use substitution. Let \( u = 3t + 2 \). Then, differentiate with respect to \( t \) to find \( du \): \(\frac{du}{dt} = 3 \quad \Rightarrow \quad dt = \frac{1}{3} du \). Now, solve for \( t \) in terms of \( u \): \( t = \frac{u - 2}{3} \).

Substitute back into the integral

Substitute \( t = \frac{u - 2}{3} \) and \( dt = \frac{1}{3} du \) back into the integral:\[ \int t (3t + 2)^{3/2} \, dt = \int \left( \frac{u - 2}{3} \right) u^{3/2} \left( \frac{1}{3} \right) \, du \]This simplifies to:\[ \frac{1}{9} \int (u - 2) u^{3/2} \, du \]

Distribute and simplify the integrand

Distribute \( u^{3/2} \) inside the integrand:\[ \frac{1}{9} \int \, (u^{5/2} - 2u^{3/2}) \, du \]Now you have:\[ \frac{1}{9} \left( \int u^{5/2} \, du - 2 \int u^{3/2} \, du \right) \]

Integrate each term

Integrate \( u^{5/2} \) and \( u^{3/2} \):\[ \int u^{5/2} \, du = \frac{2}{7} u^{7/2} + C_1 \]\[ \int u^{3/2} \, du = \frac{2}{5} u^{5/2} + C_2 \]

Combine the integrals

Substitute the results of the integration back into the expression:\[ \frac{1}{9} \left( \frac{2}{7} u^{7/2} - 2 \times \frac{2}{5} u^{5/2} \right) \]Simplify to get:\[ \frac{2}{63} u^{7/2} - \frac{4}{45} u^{5/2} + C \]

Substitute back the original variable

Replace \( u \) with \( 3t + 2 \) to write the final answer in terms of \( t \):\[ \frac{2}{63} (3t + 2)^{7/2} - \frac{4}{45} (3t + 2)^{5/2} + C \]

Key concepts

u-Substitution

U-substitution is a vital technique in integration used to simplify complex integrals. It is particularly powerful when dealing with composite functions and helps transform a problem into an easier one. The goal is to replace a part of the original integral with a single variable, usually denoted as \( u \), which turns the function into a simpler format.

Here’s how it works:

• Identify a function within the integral that can be substituted with \( u \). In this example, \( 3t + 2 \) was chosen as \( u \).
• Differentiate \( u \) with respect to the variable in the original function to find \( du \). This means if \( u = 3t + 2 \), then \( du/dt = 3 \), which implies \( dt = du/3 \).
• Substitute all instances of \( t \) and \( dt \) in the original integral with \( u \) and \( du \).

By making these substitutions, tricky integrals become significantly more manageable and solvable.

Definite Integrals

Definite integrals, unlike indefinite integrals, are used to calculate the exact area under a curve between two specified points. They are expressed with limits of integration, such as \( \int_{a}^{b} f(x) \, dx \). The solution of a definite integral is a finite value, providing precise information about the function over a given interval.

Definite integrals can be approached using u-substitution by also changing the limits of integration based on \( u \). If you change variables in the course of integration, you must also change the limits to match in terms of \( u \). It's crucial to ensure the limits align with the new variable to accurately compute the exact area, which becomes significant in physical applications like determining work or displacement.

Indefinite Integrals

Indefinite integrals are primarily used to find the antiderivative of a function. This means solving a function such as \( \int f(t) \, dt \). The solution to an indefinite integral is another function plus a constant \( C \), representing all possible antiderivatives of the original function.

Unlike definite integrals, indefinite integrals do not include limits of integration because they represent a general solution. For instance, in our exercise, using u-substitution helped simplify the complex integral \( \int t(3t + 2)^{3/2} \, dt \) to a form that’s easier to integrate.
This results in separate antiderivative terms for every component inside the integral, indicating the overall general form of all possible solutions.

Calculus Problem-Solving

Solving calculus problems effectively requires a strong grasp of techniques like u-substitution and a methodical approach to integration. Calculus is all about understanding how small changes accumulate to influence larger outcomes, such as curves and surfaces.

Effective calculus problem-solving involves:

• Spotting patterns: Carefully observing functions to identify parts suitable for substitution.
• Performing accurate algebraic manipulations: Ensuring each substitution or transformation strictly follows mathematical rules.
• Practicing multiple problems: Exposure to a variety of problems helps refine technique and understanding.

By mastering these methods, students can simplify a wide array of complex calculus problems, making the seemingly daunting task of managing mathematical expressions far more approachable.