Problem 102 - Calculus and its applications

Question

Evaluate. $$ \int \frac{t^{2}}{\sqrt[4]{2+t^{3}}} d t $$

Step-by-Step Solution

Answer

The integral is \( \frac{4}{9} (2 + t^3)^{3/4} + C \).

Step 1: Identify the Substitution

The integral looks complex due to the presence of a nested radical in the denominator. Choose a substitution to simplify. Let \( u = 2 + t^3 \), which implies \( du = 3t^2 \, dt \). Rearrange to get \( dt = \frac{du}{3t^2} \). This substitution will help simplify the integral.

Step 2: Substitute and Simplify

Substitute \( u = 2 + t^3 \) into the integral. Your expression becomes: \( \ \int \frac{t^2}{\sqrt[4]{u}} \cdot \frac{du}{3t^2} \ Which simplifies to: \ \frac{1}{3} \int \frac{1}{u^{1/4}} \, du \). The \( t^2 \) terms cancel out, leaving a simpler integral to evaluate.

Step 3: Evaluate the Simplified Integral

Now, integrate \( \int u^{-1/4} \, du \). Use the power rule for integration, which gives you: \ \frac{1}{3} \left( \frac{u^{3/4}}{3/4} \right) + C = \frac{1}{3} \times \frac{4}{3} u^{3/4} + C \ = \frac{4}{9} u^{3/4} + C \ Where \( C \) is the constant of integration.

Step 4: Back Substitute for t

Now replace \( u \) with \( 2 + t^3 \) to express the result in terms of \( t \). The final solution is: \ \frac{4}{9} (2 + t^3)^{3/4} + C.

Key Concepts

Substitution Method

The substitution method is a powerful technique used in calculus to simplify complex integrals. It involves replacing a troublesome part of an integral with a single variable, making the integration process easier. In this exercise, we used the substitution method to handle the complex expression in the integral. By letting \( u = 2 + t^3 \), the messy term \( \sqrt[4]{2+t^3} \) is replaced with a simpler expression, \( \sqrt[4]{u} \). The derivative \( du = 3t^2 \, dt \) is also calculated to express \( dt \) in terms of \( du \), easing the substitution process.

• Identify a substitution that simplifies the integral.
• Express \( dt \) in terms of the new variable \( du \).
• Replace the original variables in the integral with the substitution variables.

This method is particularly useful when dealing with integrals involving nested expressions or composite functions. Through substitution, we can transform a daunting integral into one that is more straightforward to solve.

Integration by Parts

Integration by parts is another essential method in calculus for integrating products of functions. Although it wasn't used in our specific exercise, it's a valuable technique where substitution alone doesn't simplify an integral sufficiently. The formula for integration by parts is derived from the product rule for differentiation:\[\int u \, dv = uv - \int v \, du\]To apply this method:

• Choose \( u \) and \( dv \), then differentiate \( u \) to get \( du \) and integrate \( dv \) to find \( v \).
• Substitute into the formula and solve the resulting integral \( \int v \, du \).
• This process often repeats for the \( \int v \, du \) component until a solution is reached.

Despite its absence in the original exercise, integration by parts is crucial in calculus for tackling integrals involving polynomials, logarithms, and exponential functions combined with trigonometric functions.

Definite Integrals Evaluation

Evaluating definite integrals involves finding the exact area under the curve of a function within a specified range. Unlike indefinite integrals that include a constant of integration, definite integrals provide a numerical value. To evaluate a definite integral,

• First, find the antiderivative of the function.
• Then, apply the Fundamental Theorem of Calculus, which states that if \( F(x) \) is an antiderivative of \( f(x) \), \, then \[ \int_a^b f(x) \, dx = F(b) - F(a) \]
• Substitute the upper and lower limits of the integral into the antiderivative and calculate the difference.

The approach for definite integrals is similar to indefinite integrals, but with the added step of evaluating at specific bounds. This calculation requires precision to avoid errors, especially in cases involving substitution. Our exercise involved an indefinite integral, focusing on simplifying expressions rather than calculating a specific area.

Mathematical Techniques in Calculus

In calculus, multiple mathematical techniques exist for solving integrals and derivatives, enhancing both theoretical understanding and practical problem-solving skills. This exercise primarily involved the substitution method, a common technique, but other pieces of calculus toolkit also include differentiation, reverse chain rule, integration by parts, and others.

• Each technique is chosen based on the structure and composition of the function to be integrated or differentiated.
• Substitution and integration by parts often work together to break complex functions into manageable parts.
• Differentiation and integration are inverse processes, and understanding this relationship is crucial.

Through practice, students learn to recognize which mathematical technique suits a given problem, leading to more efficient and accurate solutions. Mastering these techniques not only helps solve textbook exercises but also equips learners with tools for tackling real-world applications.