Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int t \sqrt[3]{2 t+7} d t $$

Short answer

The integral is \( t \cdot \frac{3}{8} (2t+7)^{4/3} - \frac{9}{112} (2t+7)^{7/3} + C \).

Step-by-step solution

Identify Functions for Integration by Parts

In the integration by parts formula, \( \int u \, dv = uv - \int v \, du \), we must decide which parts of the integral to set as \( u \) and \( dv \). For the integral \( \int t \sqrt[3]{2t+7} \, dt \), let's choose \( u = t \) and \( dv = \sqrt[3]{2t+7} \, dt \).

Differentiate and Integrate Functions

Differentiate \( u = t \) to find \( du \). The differential \( du = dt \). Now, integrate \( dv = \sqrt[3]{2t+7} \, dt \) to find \( v \). This involves a substitution: let \( x = 2t+7 \), then \( dx = 2 \, dt \) or \( dt = \frac{1}{2} dx \). Then the integral becomes \( \int x^{1/3} \frac{1}{2} \, dx \), which integrates to \( \frac{3}{8} x^{4/3} + C = \frac{3}{8} (2t+7)^{4/3} + C \).

Apply Integration by Parts

Substitute the equations for \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \( \int t \sqrt[3]{2t+7} \, dt \approx t \cdot \frac{3}{8} (2t+7)^{4/3} - \int \frac{3}{8} (2t+7)^{4/3} \, dt \).

Simplify and Integrate Remaining Integral

Since the remaining integral is similar to \( v \), it integrates similarly: \( \int (2t+7)^{4/3} \, dt = \frac{3}{8} (2t+7)^{7/3} \cdot \frac{1}{14} = \frac{3}{112} (2t+7)^{7/3} + C \).

Combine Results

Combine the results from the integration by parts: \( t \cdot \frac{3}{8} (2t+7)^{4/3} - \frac{9}{112} (2t+7)^{7/3} + C \).

Key concepts

Advanced Calculus

Advanced Calculus is a branch of mathematics that serves as a foundation for understanding more complex calculus concepts, such as integration by parts. Integration by parts itself is a technique used to solve more challenging integrals, especially when the integrand is a product of functions. This method originates from the product rule for differentiation, which states that the derivative of two multiplied functions can be expressed as a sum of derivatives.
Advanced Calculus enables students to deal with various methods of integration that are not straightforward, often transforming complex expressions into easier ones. In this particular exercise, the integration by parts formula is pivotal since direct integration of the given function, without this technique, would be cumbersome.

• Integration by Parts Formula: \( \int u \, dv = uv - \int v \, du \)
• Choosing appropriate \( u \) and \( dv \) is crucial for simplifying the problem.
• This method helps break down the problem into manageable steps.

The choice of \( u = t \) and \( dv = \sqrt[3]{2t+7} \, dt \) was strategic here to simplify the integral through differentiation and integration of the selected parts.

Integral Calculation

Integral Calculation involves computing the value of integrals, which is a fundamental concept in calculus used to find areas under curves and solve real-world problems like finding the total accumulated value over an interval. Here, the focus is on the actual mechanics of integration by parts.
The integral \( \int t \sqrt[3]{2t+7} \, dt \) required several transformations to reach the final answer. The integral was broken down by assigning the function parts to the variables \( u \) and \( dv \). This allowed the student to perform the necessary calculations step-by-step.

• Firstly, differentiate \( u = t \) to get \( du = dt \).
• Then, integrate \( dv = \sqrt[3]{2t+7} \, dt \) using substitution techniques.
• The substitution \( x = 2t + 7 \), with \( dx = 2 \, dt \), simplifies the integral into a manageable form \( \int x^{1/3} \frac{1}{2} \, dx \).
• Eventually, this integral becomes \( \frac{3}{8} (2t+7)^{4/3} + C \) after performing integration.

This approach ensures each part of the original integrand is handled correctly, leading to the accurate calculation of the integral.

Definite Integrals

Definite Integrals are used in calculus to calculate the net area under a curve between two specific points on a graph, giving a numerical result. Unlike indefinite integrals, which include a constant of integration and represent families of functions, definite integrals provide a specific value.
In this exercise, although we start with an indefinite integral, understanding definite integrals is essential for complete mastery of calculus as they help in evaluating the exact quantitative accumulation. If the original question required evaluating definite integrals, we would need an interval.

• Definite Integrals are expressed as \( \int_a^b f(x) \, dx \).
• This calculates the area under the curve \( f(x) \), between \( x = a \) and \( x = b \).
• When computed, this results in a numerical value, unlike indefinite integrals which have "\(+ C\)".

Although this specific problem doesn't require definite integral evaluation, knowing the differences prepares students for more complex calculus applications, enabling the translation from indefinite forms to numerical results over intervals.