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Chapter 4: Problem 45

Determine these indefinite integrals. $$\int\left(\frac{7}{\sqrt{x}}-\frac{2}{3} e^{5 x}-\frac{8}{x}\right) d x$$

Short Answer

Expert verified
14x^{1/2} - \frac{2}{15} e^{5x} - 8 \ln |x| + C

Step by step solution

01

Break Down the Integral

Split the integral into separate parts to make it easier to integrate. \[\int \left( \frac{7}{\sqrt{x}} - \frac{2}{3} e^{5x} - \frac{8}{x} \right) dx = \int \frac{7}{\sqrt{x}} \, dx - \int \frac{2}{3} e^{5x} \, dx - \int \frac{8}{x} \, dx\]
02

Simplify Each Part

Rewrite each integrand in a form that is easier to integrate. Use the properties of exponents and logarithms where necessary.\[\int 7x^{-1/2} \, dx - \frac{2}{3} \int e^{5x} \, dx - \int 8x^{-1} \, dx\]
03

Integrate Each Term

Find the antiderivative of each term.1. For \(\int 7x^{-1/2} \, dx\):Use the power rule for integration \[7 \int x^{-1/2} \, dx = 7 \left( \frac{x^{1/2}}{1/2} \right) = 14x^{1/2}\]2. For \(\int \frac{2}{3} e^{5x} \, dx\): Use the exponential rule for integration \[- \frac{2}{3} \int e^{5x} \, dx = - \frac{2}{3} \cdot \frac{1}{5} e^{5x} = - \frac{2}{15} e^{5x}\]3. For \(\int 8x^{-1} \, dx\): Use the logarithmic rule for integration \[- 8 \int x^{-1} \, dx = -8 \ln |x|\]
04

Combine the Results and Add the Constant of Integration

Combine all the results from the above integrations, and add the constant of integration, C.\[\int \left( \frac{7}{\sqrt{x}} - \frac{2}{3} e^{5x} - \frac{8}{x} \right) dx = 14x^{1/2} - \frac{2}{15} e^{5x} - 8 \ln |x| + C\]

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Key Concepts

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

Integration Techniques
When solving integrals, it's helpful to use various integration techniques to simplify the problem. One primary technique is splitting the integral into smaller, more manageable parts. This method helps to focus on each term individually. For the integral \(\int\left\(\frac{7}{\sqrt{x}}-\frac{2}{3} e^{5 x}-\frac{8}{x}\) d x\), we begin by separating it into three parts:

\[\int\left\frac{7}{\sqrt{x}}\-\frac{2}{3} e^{5 x}\-\frac{8}{x}\right\ dx = \int 7x^{-1/2} dx - \int \frac{2}{3} e^{5x} dx - \int 8x^{-1} dx \]

By handling each term separately, we can apply the right techniques to find their antiderivatives.
Exponential Functions
Exponential functions are a common part of integral calculus. When integrating exponential functions, the general rule is \(\int e^{kx} dx = \frac{1}{k} e^{kx}\), where k is a constant. For our example, consider the integral \(\int\frac{2}{3} e^{5x} dx\). Using the exponential rule, we find its antiderivative by multiplying by the reciprocal of the exponent's coefficient:

\(\- \frac{2}{3} \int e^{5x} dx = \frac{2}{3} \cdot \frac{1}{5} e^{5x} = - \frac{2}{15} e^{5x}\)

This step simplifies the integration process and helps in solving the complete integral.
Logarithmic Integration
Integrating functions in the form of \(\frac{1}{x}\) leads to logarithmic results. The integral of \(\frac{1}{x}\) is \(\ln|x|\). It's important to include the absolute value because the natural logarithm is only defined for positive values.

In our example, we have the integral \(\int 8x^{-1} dx\), which we can rewrite as \(\int \frac{8}{x} dx\). Using the logarithmic rule:

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Most popular questions from this chapter

Memorizing. The rate of memorizing information initially increases. Eventually, however, a maximum rate is reached, after which it begins to decrease. GRAPH CANT COPY Suppose that in another memory experiment the rate of memorizing is given by \(M^{\prime}(t)=-0.003 t^{2}+0.2 t\) where \(M^{\prime}(t)\) is the memory rate, in words per minute. How many words are memorized in the first 10 min (from \(t=0\) to \(t=10\) )?

Business: increasing total profit. Laso Industries finds that the marginal profit, in dollars, from the sale of \(x\) digital control boards is given by \(P^{\prime}(x)=2.6 x^{0.1}\) A customer orders 1200 digital control boards and later increases the order to \(1500 .\) Find the extra profit result. ing from the increase in order size.

The divorce rate in the United States is approximated by \(D(t)=100,000 e^{0.025 t}\) where \(D(t)\) is the number of divorces occurring at time \(t\) and \(t\) is the number of years measured from \(1900 .\) That is, \(t=0\) corresponds to \(1900, t=98 \frac{9}{365}\) corresponds to January \(9,1998,\) and so on. a) Find the total number of divorces from 1900 to \(2005 .\) Note that this is given by \(\int_{0}^{105} D(t) d t\) b) Find the total number of divorces from 1980 to 2006. Note that this is given by \(\int_{80}^{106} \mathrm{D}(t) d t\)

Evaluate. Assume \(u>0\) when ln u appears. $$\begin{array}{l} \int \frac{t^{2}+2 t}{(t+1)^{2}} d t \\ \,\left(\text { Hint: } \frac{t^{2}+2 t}{(t+1)^{2}}=\frac{t^{2}+2 t+1-1}{t^{2}+2 t+1}=1-\frac{1}{(t+1)^{2}}\right) \end{array}$$

Find the error in each of the following. Explain. $$\begin{aligned} \int_{1}^{2}\left(\ln x-e^{x}\right) d x &=\left[\frac{1}{x}-e^{x}\right]_{1}^{2} \\\&=\left(\frac{1}{2}-e^{2}\right)-\left(1-e^{1}\right) \\\&=e-e^{2}-\frac{1}{2}\end{aligned}$$
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