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Chapter 5: Problem 59

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

Short Answer

Expert verified
Question: Evaluate the following definite integral: $$\int_{1}^{2} \frac{z^{2}+4}{z} dz$$ Answer: The result of the definite integral is: $$\frac{3}{2} + 4 \ln2$$

Step by step solution

01

Rewrite the integrand

First, let's rewrite the integrand as two separate fractions $$\int_{1}^{2} \frac{z^{2}+4}{z} dz = \int_{1}^{2} (\frac{z^2}{z} + \frac{4}{z}) dz$$
02

Simplify the integrand

Now we can simplify the expression: $$\int_{1}^{2} (\frac{z^2}{z} + \frac{4}{z}) dz = \int_{1}^{2} (z + \frac{4}{z}) dz$$
03

Find the antiderivative F(z) using the power rule

We find the antiderivative of each term. For z, we use the power rule: raise the power by 1 and divide by the new power. For \(\frac{4}{z}\), recognize that \(\frac{1}{z} = z^{-1}\) and then apply the power rule. $$F(z) = \int (z + \frac{4}{z}) dz = \int z^1 dz + 4 \int z^{-1} dz = \frac{1}{2}z^2 + 4 \ln|z| + C$$
04

Apply the Fundamental Theorem of Calculus

Now, using the Fundamental Theorem of Calculus, we can directly evaluate the definite integral by substituting the bounds of the integral (1 and 2) into the antiderivative function. $$\int_{1}^{2} (z + \frac{4}{z}) dz = F(2) - F(1)$$
05

Evaluate F(2) and F(1)

$$F(2) = \frac{1}{2}(2)^2 + 4 \ln|2| = 2 + 4 \ln2$$ $$F(1) = \frac{1}{2}(1)^2 + 4 \ln|1| = \frac{1}{2} + 4(0) = \frac{1}{2}$$
06

Calculate the definite integral

Now we subtract F(1) from F(2): $$\int_{1}^{2} (z + \frac{4}{z}) dz = (2 + 4 \ln2) - \frac{1}{2} = \frac{3}{2} + 4 \ln2$$ So, the result of the definite integral is: $$\boxed{\frac{3}{2} + 4 \ln2}$$

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Key Concepts

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

Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a bridge between differential and integral calculus. It connects the concept of differentiating a function with integrating a function. Specifically, it states that if a function is continuous on a closed interval, then the function has an antiderivative on that interval.
  • The first part of the theorem states that if you have a continuous function and you integrate it, then differentiate the result, you end up back at the original function.
  • The second part of the theorem tells us that if you have an antiderivative of a function, you can evaluate a definite integral by plugging in the limits of integration into the antiderivative.
In our exercise, to evaluate \(\int_{1}^{2} \frac{z^{2}+4}{z} dz\) using this theorem, we first needed to find the antiderivative and then substitute the upper and lower limits of integration.
Antiderivative
An antiderivative, also known as an indefinite integral, is a function whose derivative gives back the original function. In our problem, we found the antiderivative for \(z + \frac{4}{z}\) by breaking it into manageable pieces.
  • For the term \(z\), the antiderivative is calculated using the power rule in reverse.
  • For \(\frac{4}{z} = 4z^{-1}\), the antiderivative becomes 4 \ln|z|\.
Therefore, the antiderivative in this case is given by \(\frac{1}{2}z^2 + 4 \ln|z| + C\). This function represents a family of functions, each shifted vertically by a constant \(C\). But for definite integrals, the constant cancels out during evaluation.
Definite Integral Evaluation
Definite integral evaluation involves calculating the exact area under the curve of a function between two points. Once you have the antiderivative, the Fundamental Theorem of Calculus allows you to evaluate this integral by calculating: \[F(b) - F(a)\]where \(F(x)\) is the antiderivative of the function being integrated. In the original exercise solution, we did this using the antiderivatives calculated:
  • First, substitute the upper bound \(2\) into \(F(z)\) to get \(F(2) = 2 + 4 \ln2\).
  • Then substitute the lower bound \(1\) yielding \(F(1) = \frac{1}{2}\). The natural log term goes to zero since \(\ln|1|=0\).
Subtracting these results gives the area under the curve, which is \(\frac{3}{2} + 4 \ln2\).
Power Rule in Integration
The Power Rule in Integration is a simple yet powerful tool used to find antiderivatives. This rule is essentially the reverse of the power rule for differentiation. To apply it, if the function is of the form \(x^n\), then its antiderivative is \[\frac{x^{n+1}}{n+1} \, + \, C\]where \(n eq -1\). In our exercise, the power rule was applied to \(z^1\), resulting in \[\frac{z^{2}}{2}\]For \(\frac{4}{z}\), recognizing it as \(4z^{-1}\), the special case for \(n = -1\) applies, leading to the antiderivative \(4 \ln|z|\). This illustrates how the power rule simplifies finding antiderivatives and evaluating integrals.

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Most popular questions from this chapter

The family of parabolas \(y=\frac{1}{a}-\frac{x^{2}}{a^{3}}\) where \(a>0,\) has the property that for \(x \geq 0,\) the \(x\) -intercept is \((a, 0)\) and the \(y\) -intercept is \((0,1 / a) .\) Let \(A(a)\) be the area of the region in the first quadrant bounded by the parabola and the \(x\) -axis. Find \(A(a)\) and determine whether it is an increasing, decreasing, or constant function of \(a\)

Shape of the graph for right Riemann sums Suppose a right Riemann sum is used to approximate the area of the region bounded by the graph of a positive function and the \(x\) -axis on the interval \([a, b] .\) Fill in the following table to indicate whether the resulting approximation underestimates or overestimates the exact area in the four cases shown. Use a sketch to explain your reasoning in each case. $$\begin{array}{|l|l|l|}\hline & \text { Increasing on }[a, b] & \text { Decreasing on }[a, b] \\\\\hline \text { Concave up on }[a, b] & & \\\\\hline \text { Concave down on }[a, b] & & \\\\\hline\end{array}$$

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals. $$\int_{0}^{2} \frac{2 x}{\left(x^{2}+1\right)^{2}} d x$$

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating. $$\int \frac{3}{\sqrt{1-25 x^{2}}} d x$$

Approximating areas Estimate the area of the region bounded by the graph of \(f(x)=x^{2}+2\) and the \(x\)-axis on [0,2] in the following ways. a. Divide [0,2] into \(n=4\) subintervals and approximate the area of the region using a left Riemann sum. Illustrate the solution geometrically. b. Divide [0,2] into \(n=4\) subintervals and approximate the area of the region using a midpoint Riemann sum. Illustrate the solution geometrically. c. Divide [0,2] into \(n=4\) subintervals and approximate the area of the region using a right Riemann sum. Illustrate the solution geometrically.
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