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Chapter 6: Problem 37

Evaluate the definite integral. $$ \int_{0}^{1} \frac{x^{3}}{x^{2}-2} d x $$

Short Answer

Expert verified
The value of the definite integral is 0.5.

Step by step solution

01

Substitution

The strategy here is to apply a substitution. Let \( u = x^{2} - 2 \). Then \( du = 2x dx \). It follows that \( x dx = \frac{{du}}{2} \). Additionally, when \( x = 0 \), \( u = -2 \), and when \( x = 1 \), \( u = -1 \). The integral becomes \(-\frac{1}{2}\int_{-2}^{-1} \frac{u du}{u} \).
02

Simplify the Integral

After the substitution, the integral simplifies to a polynomial, which is easier to solve. The previous expression simplifies further to \(-\frac{1}{2}\int_{-2}^{-1} du \).
03

Compute the Integral

We compute the integral: \(-\frac{1}{2}[u]_{-2}^{-1} = \frac{1}{2} = 0.5 \).

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Key Concepts

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

Substitution Method
The substitution method is a technique used to simplify complex integrals by transforming them into a more manageable form. In our example, we started with \[ \int_{0}^{1} \frac{x^{3}}{x^{2}-2} \ dx \].
To simplify, we chose a new variable, \( u \), such that \( u = x^2 - 2 \). This substitution simplifies the integrand significantly.
  • Substitute \( x \) with \( u \) using \( u = x^2 - 2 \).
  • Calculate the differential \( du \). Here, \( du = 2x \, dx \), so \( x \, dx = \frac{du}{2} \).
  • Adjust the limits of the integral to match the new variable. When \( x = 0 \), \( u = -2 \), and when \( x = 1 \), \( u = -1 \).
This conversion made the integral simpler, as the new form is \[-\frac{1}{2} \int_{-2}^{-1} \frac{u \, du}{u}.\]
Being comfortable with changing variables in integrals is crucial as it helps make seemingly complex problems much easier to tackle.
Integral Simplification
Integral simplification is the process of reducing an integral to a simpler form that is easier to evaluate. After substitution, our complex integral became:
\[-\frac{1}{2} \int_{-2}^{-1} \frac{u \, du}{u}.\]
At this stage, cancelling out common terms is essential.
  • Here, the \( u \) terms in \( \frac{u}{u} \) canceled each other, leaving just \(-\frac{1}{2} \int_{-2}^{-1} du \).
Such simplifications help reduce the computational effort required to solve the integral.
Once the expression is simplified, we're left with evaluating a basic integral, \[\int_{-2}^{-1} du,\] which is straightforward. This step exhibits the power of simplification, allowing us to focus more on integral evaluation than algebraic manipulation.
Polynomial Integration
Polynomial integration is one of the foundational techniques in calculus. It's often straightforward and results from integrating polynomial expressions directly. In our problem's final stage, successful use of substitution and simplification led to:
\[-\frac{1}{2} \int_{-2}^{-1} du.\]
This integral, being a polynomial, is directly solvable. We integrate the constant term with respect to \( u \), resulting in \(-\frac{1}{2}[u]_{-2}^{-1}.\)
  • The integration of 1 with respect to \( u \) is \( u \).
  • Apply the definite integral limits: substitute \( u = -1 \) and \( u = -2 \) in the result and compute the difference.
  • This yields \(-\frac{1}{2} ((-1) - (-2)) = \frac{1}{2}.\)
Through these steps, polynomial integration makes the process of finding areas under linear and polynomial curves manageable and efficient.

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Most popular questions from this chapter

Decide whether the integral is improper. Explain your reasoning. $$ \int_{1}^{3} \frac{d x}{x^{2}} $$

Use the Trapezoidal Rule and simpson's Rule to approximate the value of the definite integral for the indicated value of \(n\). Compare these results with the exact value of the definite integral. Round your answers to four decimal places. $$ \int_{1}^{2} \frac{1}{x} d x, n=4 $$

Consider the region satisfying the inequalities. Find the area of the region. $$ y \leq e^{-x}, y \geq 0, x \geq 0 $$

Explain why the integral is improper and determine whether it diverges or converges. Evaluate the integral if it converges. $$ \int_{0}^{2} \frac{1}{(x-1)^{2}} d x $$

Capitalized cost Find the capitalized cost \(C\) of an asset (a) for \(n=5\) years, (b) for \(n=10\) years, and (c) forever. The capitalized cost is given by $$ C=C_{0}+\int_{0}^{n} c(t) e^{-r t} d t $$ where \(C_{0}\) is the original investment, \(t\) is the time in years, \(r\) is the annual interest rate compounded continuously, and \(c(t)\) is the annual cost of maintenance (measured in dollars). [Hint: For part (c), see Exercises \(35-38 .]\) $$ C_{0}=\$ 650,000, c(t)=25,000(1+0.08 t), r=12 \% $$
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