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Chapter 5: Problem 93

Evaluate the following definite integrals using the Fundamental Theorem of Calculus. $$\int_{1}^{2} \frac{z^{2}+4}{z} d z$$

Short Answer

Expert verified
Question: Evaluate the definite integral $\int_{1}^{2} \frac{z^{2}+4}{z} d z$. Answer: The definite integral evaluates to $\frac{3}{2} + 4 \ln2$.

Step by step solution

01

1. Simplify the integrand

The integrand can be simplified by dividing each term in the numerator by the denominator: $$\frac{z^{2}+4}{z} = \frac{z^2}{z} + \frac{4}{z} = z + \frac{4}{z}$$
02

2. Calculate the antiderivative

Now that the integrand is simplified, we can calculate the antiderivative: $$\int (z + \frac{4}{z}) d z = \int z d z + \int \frac{4}{z} d z$$ It's easier to find the antiderivative for each term separately. The antiderivative of \(z\) is \(\frac{z^2}{2}\), and the antiderivative of \(\frac{4}{z}\) is \(4 \ln|z|\). We add the constant C: $$\frac{z^2}{2} + 4 \ln|z| + C $$
03

3. Apply the Fundamental Theorem of Calculus

Now we can apply the Fundamental Theorem of Calculus to evaluate the definite integral: $$\int_{1}^{2} (z +\frac{4}{z}) d z= \left[\frac{z^2}{2} + 4 \ln|z|\right]_{1}^{2} = \left(\frac{2^2}{2} + 4 \ln|2|\right) - \left(\frac{1^2}{2} + 4 \ln|1|\right)$$ $$= \left(2 + 4 \ln2\right) - \left(\frac{1}{2} + 0\right) = \frac{3}{2} + 4 \ln2$$ Therefore, the answer to the definite integral is: $$\int_{1}^{2} \frac{z^{2}+4}{z} d z = \frac{3}{2} + 4 \ln2$$

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