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Use the method of substitution to evaluate the definite integrals. $$ \int_{-1}^{0} 24 \frac{x^{2}}{\left(x^{3}-1\right)^{5}} d x $$

Short Answer

Expert verified
The value of the definite integral is \(-\frac{15}{8}\).

Step by step solution

01

Choose a Substitution

Choose a substitution to simplify the integral. Let's set \( u = x^3 - 1 \). This choice is inspired by the expression in the denominator \((x^3 - 1)^5\).
02

Differentiate the Substitution

Find \( du \) in terms of \( dx \). Differentiate \( u = x^3 - 1 \) to get \( du = 3x^2 \, dx \).
03

Solve for dx

We need \( dx \) in terms of \( du \) and other variables involved. From \( du = 3x^2 \, dx \), we solve for \( dx \) as \( dx = \frac{du}{3x^2} \).
04

Substitute in the Integral

Replace \( x^2 \, dx \) in the integral with \( \frac{du}{3} \) using \( x^2 = \frac{1}{3} du \). The integral transforms to:\[ \int_{u(-1)}^{u(0)} 24 \frac{1}{u^5} \cdot \frac{du}{3} \] Simplify this to:\[ \int_{0}^{-1} 8u^{-5} \, du \] because \( x = -1 \rightarrow u = (-1)^3 - 1 = -2 \) and \( x = 0 \rightarrow u = (0)^3 - 1 = -1 \).
05

Evaluate the New Integral

The integral becomes:\[ 8 \int_{-2}^{-1} u^{-5} \, du \]Integrate \( u^{-5} \) to find \( \frac{-1}{4} u^{-4} \). Evaluate it from \( -2 \) to \( -1 \):\[ 8\left( \frac{-1}{4}(-1)^{-4} - \frac{-1}{4}(-2)^{-4} \right) \].
06

Calculate the Definite Integral

Plug in the limits of integration to get:\[ 8\left( \frac{-1}{4}(1) - \frac{-1}{4}\left(\frac{1}{16}\right) \right) = 8 \left( -\frac{1}{4} + \frac{1}{64} \right) \].Simplify it to\[ 8\left( -\frac{16}{64} + \frac{1}{64} \right) = 8 \cdot \left( -\frac{15}{64} \right) = -\frac{120}{64} = -\frac{15}{8} \].
07

Simplify the Final Result

The result of the evaluation simplifies to \(-\frac{15}{8}\). This is the value of the definite integral.

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Key Concepts

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

Definite Integrals
Definite integrals are a vital part of calculus, helping us find the exact area under a curve between two points. Unlike indefinite integrals, which produce a family of functions, definite integrals have specific numerical values. This operation is key to many applications in physics, engineering, and beyond. For a definite integral, we have:
  • The integrand: the function you're integrating, in our case, \( 24 \frac{x^{2}}{(x^{3}-1)^{5}} \).
  • The limits of integration: the endpoints of the interval, here from \( -1 \) to \( 0 \).
  • The variable of integration, \( dx \).

When we evaluate a definite integral, we aren't just finding an antiderivative; we're calculating the exact area, taking into account both positive and negative contributions based on the function's position relative to the x-axis.
The application of definite integrals can involve substituting a simpler expression as we did, substantially transforming the problem into a more manageable form.
Substitution Method
The substitution method is a powerful technique in calculus to simplify integrals. It involves changing variables to transform the integral into one that is easier to solve. In this method, you choose a new variable \( u \) to replace a part of the original integrand. Here’s how it’s done:
  • Identify a substitution: Choose \( u = x^3 - 1 \), simplifying the exponent-heavy term in our original integral.
  • Differentiate \( u \) to find \( du \): From \( u = x^3 - 1 \), differentiate to get \( du = 3x^2 \, dx \).
  • Express \( dx \) in terms of \( du \): Solving gives \( dx = \frac{du}{3x^2} \).
  • Replace \( x^2 \, dx \) in the integral: Substitution changes the integral's form, simplifying it significantly.

This method, akin to reversing the chain rule, is especially useful when dealing with integrals involving complex functions, enabling more straightforward integration.
Calculus Problems
Calculus problems often necessitate employing diverse strategies to tackle complex integrals. Solving these problems demands an understanding of several techniques and concepts. Here's a breakdown:
  • Recognizing Patterns: Identifying parts of the integral that can be substituted or simplified.
  • Calculation Techniques: In our example, we moved from a complicated expression to a simpler polynomial form.
  • Precision in Execution: Ensure all steps, like substituting limits of integration, are accurately followed to avoid errors.

Approaching calculus problems with a clear strategy can transform seemingly daunting tasks into achievable solutions. It's like solving a puzzle where each strategic move gets you closer to the final piece. By mastering substitution and integral evaluation, solving calculus problems becomes a more structured and manageable process.

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Most popular questions from this chapter

A sum of integrals of the form \(\int_{a}^{b} f(x) d x\) is given. Express the sum as a single integral of form \(\int_{c}^{d} g(y) d y\). $$ \int_{-2}^{0} \sqrt{x+2} d x+\int_{0}^{2}(\sqrt{x+2}-x) d x $$

A function \(f,\) an interval \(I,\) and an even integer \(N\) are given. Approximate the integral of \(f\) over \(I\) by partitioning \(I\) into \(N\) equal length subintervals and using the Midpoint Rule, the Trapezoidal Rule, and then Simpson's Rule. $$ f(x)=\cos (x) \quad I=[\pi / 4,5 \pi / 4], N=4 $$

A function \(f\) is defined piecewise on an interval \(I=[a, b] .\) Find the area of the region that is between the vertical lines \(x=a\) and \(x=b\) and between the graph of \(f\) and the \(x\) -axis. $$ f(x)=\left\\{\begin{array}{ll} \sin (x) & \text { if } 0 \leq x \leq \pi / 4 \\ \cos (x) & \text { if } \pi / 4

A function \(f\) is defined on a specified interval \(I=[a, b] .\) Calculate the area of the region that lies between the vertical lines \(x=a\) and \(x=b\) and between the graph of \(f\) and the \(x\) -axis. \(f(x)=2 x^{2}-8 \quad I=[-3,5]\)

The partition \\{0,15,25,50,75,90,100\\} of [0,100] given in Exercise 46 precludes an application of Simpson's Rule because it is not uniform. However, because the number of subintervals is even, the idea behind Simpson's Rule can be used. Find the degree 2 polynomial \(p_{1}\) that passes through the three points \((0,0),(15,3),\) and \((25,19) .\) The interpolation command of a computer algebra system will make short work of finding this polynomial. Calculate \(\int_{0}^{25} p_{1}(x) d x .\) Next, find the degree 2 polynomial \(p_{2}\) that passes through the three points \((25,19),(50,25),\) and \((75,42) .\) Calculate \(\int_{25}^{75} p_{2}(x) d x .\) Repeat this procedure with the last three points, \((75,42),(90,75),\) and (100,100) . What approximation of \(\int_{0}^{100} L(x) d x\) results?

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