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Chapter 23: Problem 599

Integrate: \(\int\left(\mathrm{x}^{3}-7\right)^{8} 3 \mathrm{x}^{2} \mathrm{dx}\)

Short Answer

Expert verified
The integration of \(\int (x^3 - 7)^8 \cdot 3x^2 dx\) is solved using substitution method: First, let \(u = x^3 - 7\) and then find the derivative of u with respect to x, \(du = 3x^2 dx\). Substitute the variables back into the integral, resulting in \(\int u^8 du\). Integrate with respect to u, to get \(\frac{1}{9}u^9 + C\). Finally, replace u with the original function of x, resulting in the short answer: \(\frac{1}{9}(x^3 - 7)^9 + C\).

Step by step solution

01

Identify a suitable substitution

The substitution method is most effective when the integrand has a more complex function inside a simpler one. Here, we have \((x^3 - 7)^8\), where \(x^3 - 7\) is inside the power of 8. We can choose the substitution \(u = x^3 - 7\).
02

Find du and dx

Now we need to compute the derivative of u with respect to x: \(du = \frac{d(u)}{d(x)} dx\) \(du = \frac{d(x^3 - 7)}{d(x)} dx\) \(du = (3x^2) dx\)
03

Substitute u and du into the integral

Now we can replace \(x^3 - 7\) with \(u\) in the original integral, and \(3x^2 dx\) with \(du\): \(\int (u)^8 \, du\)
04

Integrate with respect to u

Next, we will integrate the new function with respect to the variable u: \(\int u^8 \, du = \frac{1}{9}u^9 + C\)
05

Replace u with the original function of x

Finally, we will replace u with the original function of x: \(\frac{1}{9}(x^3 - 7)^9 + C\) The final answer is: \(\boxed{\frac{1}{9}(x^3 - 7)^9 + C}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is a fundamental branch of mathematics that deals with change and motion. It provides tools to study and analyze continuous change, akin to how algebra deals with static equations. Calculus is divided primarily into two branches: differential calculus and integral calculus.
  • Differential calculus focuses on the concept of the derivative, exploring rates of change and slopes of curves.
  • Integral calculus, on the other hand, is focused on accumulation of quantities and the areas under and between curves.
In problems like the given original exercise, where we integrate a function, we're engaging with integral calculus. Understanding these basics is crucial when solving calculus problems, as it gives us the ability to interpret and manipulate functions describing real-world scenarios.
Definite Integrals
Definite integrals help us calculate the total accumulation of a quantity over a given interval. They have limits of integration, represented by numbers on the integral sign and yield a specific numerical value. This value can represent numerous physical concepts, such as
  • area under a curve,
  • total distance traveled by a moving object,
  • or even accumulated growth in a biological system.
While the exercise we worked on is about indefinite integrals, having a grasp of definite integrals helps solidify your understanding of how integration can be applied in practical, quantifiable situations.
Indefinite Integrals
An indefinite integral, unlike a definite integral, does not have specified limits of integration and represents a family of functions. This family is characterized by an arbitrary constant, often denoted as "C" in the solution.
  • In the context of the original exercise, we dealt with an indefinite integral as it required finding a general antiderivative of the function.
  • The solution to this integral will include a constant of integration, which accounts for the fact that there could be multiple functions that have the same derivative.
Understanding indefinite integrals is foundational in learning about how functions operate and change across intervals without the constraint of specific limits.
u-substitution
u-substitution is a technique used in integration that simplifies the integration process. It involves choosing a substitution that lets you re-frame a complex integral into an easier one. In our exercise, we identified a suitable substitution with
  • \( u = x^3 - 7 \)
  • This conversion simplifies the expression \((x^3 - 7)^8 3x^2\) into \( u^8 du\).
By making these substitutions, we effectively transformed the integral into a form that is simpler and more straightforward to solve. It's like using a trail map that leads you through a more navigable path to reach the solution.
Polynomial Functions
Polynomial functions are expressions consisting of variables and coefficients, combined using only addition, subtraction, multiplication, and non-negative integer exponents. They are a key part of calculus as they serve as basic building blocks for more complex functions.
  • Our exercise involved polynomials within its expression, specifically \( x^3 - 7 \), which is a cubic polynomial.
  • Handling polynomials is often more straightforward due to their regular behavior and the predictable patterns of their derivatives and integrals.
Understanding how to work with polynomial functions by performing operations such as integration and differentiation is a fundamental skill in calculus, enabling exploration of more advanced mathematical concepts.

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Most popular questions from this chapter

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