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Chapter 5: Problem 37

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{1 / 2}^{1}\left(x^{-3}-8\right) d x$$

Short Answer

Expert verified
Question: Evaluate the definite integral of the function \(f(x) = x^{-3} - 8\) from \(x = \frac{1}{2}\) to \(x = 1\). Answer: The value of the definite integral is \(-\frac{5}{2}\).

Step by step solution

01

Write the general equation for the antiderivative

The general equation for an antiderivative is given by: $$F(x) = \int f(x) \, dx$$ Note that we will find the antiderivative for the function \(f(x) = x^{-3} - 8\).
02

Find the antiderivative for each term

To find the antiderivative of the given function, we treat each term separately: $$F(x) = \int (x^{-3} - 8) \, dx = \int x^{-3} \, dx - \int 8 \, dx$$
03

Antiderivative of \(x^{-3}\)

To find the antiderivative of \(x^{-3}\), we use the power rule: $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$ So, the antiderivative of \(x^{-3}\) is: $$\int x^{-3} \, dx = \frac{x^{-2}}{-2} + C_1$$
04

Antiderivative of \(8\)

The antiderivative of a constant is equal to the constant times x: $$\int 8 \, dx = 8x + C_2$$
05

Combine the antiderivatives

Now we combine the antiderivatives of the two terms to find the antiderivative of \(f(x)\): $$F(x) = \frac{x^{-2}}{-2} + C_1 - 8x - C_2$$ #Step 2: Evaluate the antiderivative at the limits of integration#
06

Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that: $$\int_a^b f(x) \, dx = F(b) - F(a)$$ In this case, \(a = \frac{1}{2}\) and \(b = 1\). We will evaluate the antiderivative, \(F(x)\), at these limits.
07

Evaluate \(F(1)\)

Substituting \(x = 1\) into the antiderivative, we get: $$F(1) = \frac{1^{-2}}{-2} + C_1 - 8(1) - C_2 = \frac{1}{-2} + C_1 - 8 - C_2$$
08

Evaluate \(F(\frac{1}{2})\)

Substituting \(x = \frac{1}{2}\) into the antiderivative, we get: $$F\left(\frac{1}{2}\right) = \frac{\left(\frac{1}{2}\right)^{-2}}{-2} + C_1 - 8\left(\frac{1}{2}\right) - C_2 = \frac{4}{-2} + C_1 - 4 - C_2 = -2 + C_1 - 4 - C_2$$
09

Evaluate the definite integral

Using the Fundamental Theorem of Calculus and the values of \(F(1)\) and \(F(\frac{1}{2})\), we find the value of the definite integral: $$\int_{1/2}^1 (x^{-3} - 8) \, dx = F(1) - F\left(\frac{1}{2}\right) = \left(\frac{1}{-2} + C_1 - 8 - C_2\right) - \left(-2 + C_1 - 4 - C_2\right) = \frac{1}{-2} + 2 - 8 + 4 = -\frac{1}{2} - 2$$ So, the value of the definite integral is: $$\int_{1/2}^1 (x^{-3} - 8) \, dx = -\frac{5}{2}$$

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