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Chapter 5: Problem 41

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{-2}^{-1} x^{-3} d x$$

Short Answer

Expert verified
Question: Evaluate the definite integral $$\int_{-2}^{-1} x^{-3} dx$$ Answer: $$-\frac{3}{8}$$

Step by step solution

01

Finding the Antiderivative

To find the antiderivative of \(x^{-3}\), we need to apply the power rule: If \(n\neq-1\), then: $$\int x^n dx = \frac{x^{n+1}}{n+1}+ C$$ For \(x^{-3}\), we have \(n = -3\), so we can apply the power rule: $$\int x^{-3} dx = \frac{x^{(-3+1)}}{(-3+1)} + C = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C$$
02

Using the Fundamental Theorem of Calculus (Part 2)

Now, we will use the Fundamental Theorem of Calculus to evaluate the definite integral: $$\int_{-2}^{-1} x^{-3} dx = \left[-\frac{1}{2x^2}\right]_{-2}^{-1}$$ This means we need to evaluate the antiderivative at \(x = -1\) and \(x = -2\), and then find the difference between these values.
03

Evaluating the Antiderivative at the Limits of Integration

First, we will evaluate the antiderivative at \(x = -1\): $$-\frac{1}{2(-1)^2} = -\frac{1}{2(1)} = -\frac{1}{2}$$ Next, we will evaluate the antiderivative at \(x = -2\): $$-\frac{1}{2(-2)^2} = -\frac{1}{2(4)} = -\frac{1}{8}$$
04

Finding the Difference between the Antiderivative at the Limits of Integration

Now, we will find the difference between the antiderivative at the upper limit of integration and the lower limit of integration: $$\left[-\frac{1}{2x^2}\right]_{-2}^{-1} = \left(-\frac{1}{2}\right) - \left(-\frac{1}{8}\right) = -\frac{1}{2} + \frac{1}{8} = -\frac{3}{8}$$ Thus, the value of the integral is: $$\int_{-2}^{-1} x^{-3} dx = -\frac{3}{8}$$

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Most popular questions from this chapter

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