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Chapter 5: Problem 21

Evaluate the integral and check your answer by differentiating. $$\int\left[\frac{2}{x}+3 e^{x}\right] d x$$

Short Answer

Expert verified
The integral is \( 2\ln|x| + 3e^x + C \). Verified by differentiation.

Step by step solution

01

Split the Integral

The given integral can be split into two separate integrals. This gives us: \[ \int \left[ \frac{2}{x} + 3e^x \right] \, dx = \int \frac{2}{x} \, dx + \int 3e^x \, dx \]
02

Integrate Each Term Separately

First, we integrate \( \int \frac{2}{x} \, dx \) which is \( 2\int \frac{1}{x} \, dx = 2\ln|x| + C \). Next, integrate \( \int 3e^x \, dx \) which results in \( 3e^x + C \).
03

Combine the Results

Combine the results from the two integrals to form the complete solution: \[ \int \left[ \frac{2}{x} + 3e^x \right] \, dx = 2\ln|x| + 3e^x + C \] where \( C \) is the constant of integration.
04

Differentiate to Check Your Solution

To verify our integral, differentiate the result \( 2\ln|x| + 3e^x + C \). The derivative of \( 2\ln|x| \) is \( \frac{2}{x} \) and the derivative of \( 3e^x \) is \( 3e^x \). The constant \( C \) becomes zero upon differentiation. Thus, \[ \frac{d}{dx}(2\ln|x| + 3e^x + C) = \frac{2}{x} + 3e^x \]. This matches the original function, confirming our integration is correct.

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Key Concepts

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

Differentiation Check
When solving integrals, a powerful tool for checking our work is performing a differentiation check. After finding the integral solution, differentiating it shows if we get back the original integrand. If what you differentiate matches the given function you integrated, it means your integration was accurate.

Here's a quick recap of the differentiation process for verification:
  • Take the derivative of the entire resulting expression from your integration.
  • Ensure every term matches the original components in the integrand.
For example, if your integrated result is \(2\ln|x| + 3e^x + C\), differentiating should yield \( \frac{2}{x} + 3e^x\). The constant of integration (\

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Most popular questions from this chapter

Determine whether the statement is true or false. Explain your answer. If \(f(x)\) is continuous on the interval \([a, b],\) and if the definite integral of \(f(x)\) over this interval has value 0 , then the equation \(f(x)=0\) has at least one solution in the interval \([a, b]\)

(a) Evaluate \(\int\left[x /\left(x^{2}+1\right)\right] d x\) (b) Use a graphing utility to generate some typical integral curves of \(f(x)=x /\left(x^{2}+1\right)\) over the interval (-5,5)

$$\begin{aligned} \sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=&\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right] \\\ &+\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right] \\ =& 5^{3}-1^{3}=124 \end{aligned}$$ For convenience, the terms are listed in reverse order. Note how cancellation allows the entire sum to collapse like a telescope. A sum is said to telescope when part of each term cancels part of an adjacent term, leaving only portions of the first and last terms uncanceled. Evaluate the telescoping sums in these exercises. $$\sum_{k=5}^{17}\left(3^{k}-3^{k-1}\right)$$

(a) Show that $$\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\dots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}$$ $$\left[\text {Hint: } \frac{1}{(2 n-1)(2 n+1)}=\frac{1}{2}\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right)\right]$$ (b) Use the result in part (a) to find $$\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{1}{(2 k-1)(2 k+1)}$$

$$\begin{aligned} \sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=&\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right] \\\ &+\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right] \\ =& 5^{3}-1^{3}=124 \end{aligned}$$ For convenience, the terms are listed in reverse order. Note how cancellation allows the entire sum to collapse like a telescope. A sum is said to telescope when part of each term cancels part of an adjacent term, leaving only portions of the first and last terms uncanceled. Evaluate the telescoping sums in these exercises. $$\sum_{k=1}^{50}\left(\frac{1}{k}-\frac{1}{k+1}\right)$$
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