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Chapter 4: Problem 8

Find the general antiderivative. $$\int\left(2 x^{-2}+\frac{1}{\sqrt{x}}\right) d x$$

Short Answer

Expert verified
The antiderivative of the given function is \(-\frac{2}{x} + 2 \sqrt{x} + C\).

Step by step solution

01

Rewrite the function

First, rewrite the integral into two separate integrals and simplify \(\sqrt{x}\) to \(x^{1/2}\), and \(x^{-2}\) inside the integral to \(x^{2}\) in the denominator. So we get: \[\int(2 x^{-2}+\frac{1}{\sqrt{x}}) dx = \int 2x^{-2} dx + \int x^{-1/2} dx.\]
02

Calculate the integral

The anti-derivative of \(x^n\) is \(\frac{x^{n+1}}{n+1}\). Therefore, for the first integral the power is -2 and for the second one, it's -1/2. Now, apply the power rule to get: \[-2x^{-1} + 2x^{1/2} + C\]. Here, C is the constant of integration.
03

Simplify the solution

Finally, simplify the result. In order to get the final solution rewrite the terms to the more readable form : \[-\frac{2}{x} + 2 \sqrt{x} + C. \]

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Key Concepts

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

Integration
Integration is a fundamental concept in calculus that is often used to find the area under a curve, among other applications. At its core, it is the reverse process of differentiation. While differentiation involves finding the rate of change of a function, integration is concerned with finding the original function given this rate of change, or the 'accumulated quantity'. Think of it like piecing together the total distance traveled from the speed at any given point in time. In the context of our exercise, integration helps us find the general antiderivative of a given function.
Power Rule for Integration
The power rule for integration is a technique used to find the antiderivative of functions in the form of x^n. According to this rule, the antiderivative of x^n is (x^(n+1))/(n+1), provided that n is not equal to -1. This rule simplifies the process of integration when dealing with polynomial functions or terms. In our exercise, we applied the power rule to the terms 2x^{-2} and x^{-1/2}, by increasing the exponents by one and dividing by the new exponent, which effectively 'reverses' the process of differentiation for these terms.
Constant of Integration
The constant of integration, represented by the symbol C, is an essential part of the general antiderivative. When we find the antiderivative of a function, it's important to remember that there could be an infinite number of functions that differentiate back to the original function; these functions differ by a constant. Therefore, we include C to account for all possible functions that share the same derivative. It represents the unknown, or arbitrary, constant that would be lost during the differentiation of the original function.
Antiderivative Calculation
The calculation of an antiderivative involves applying rules of integration to find a function whose derivative is the original function given. In our exercise, after rewriting the integral and applying the power rule, we find separate antiderivatives for each term. The process of combining these results and adding the constant of integration gives us the general antiderivative of the entire function. It's a systematic approach to reversing differentiation, and it's crucial to recognize how terms are manipulated to fit the forms for which we have integration rules, such as the power rule explained earlier.

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Most popular questions from this chapter

Identify the integrals to which the Fundamental Theorem of Calculus applies; the other integrals are called improper integrals. (a) \(\int_{0}^{4} \frac{1}{x-4} d x\) (b) \(\int_{0}^{1} \sqrt{x} d x\) (c) \(\int_{0}^{1} \ln x d x\)

In the text, we deferred the proof of \(\lim _{h \rightarrow \infty} \frac{e^{h}-1}{h}=1\) to the exercises. In this exercise, we guide you through one possible proof. (Another proof is given in exercise 42.) Starting with \(h>0,\) write \(h=\ln e^{h}=\int_{1}^{e^{h}} \frac{1}{x} d x .\) Use the Integral Mean Value Theorem to write \(\int_{1}^{e^{h}} \frac{1}{x} d x=\frac{e^{h}-1}{\bar{x}}\) for some number \(\bar{x}\) between 1 and \(e^{h} .\) This gives you \(\frac{e^{h}-1}{h}=\bar{x} .\) Now, take the limit as \(h \rightarrow 0^{+} .\) For \(h<0,\) repeat this argument, with \(h\) replaced with \(-h.\)

Use the Fundamental Theorem of Calculus to find an antiderivative of \(\sin \sqrt{x^{2}+1}\)

Evaluate the integral exactly, if possible. Otherwise, estimate it numerically. (a) \(\int_{0}^{\pi} \sin x^{2} d x\) (b) \(\int_{0}^{\pi} x \sin x^{2} d x\)

In this exercise, we guide you through a different proof of \(\lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=1 .\) Start with \(f(x)=\ln x\) and the fact that \(f^{\prime}(1)=1 .\) Using the alternative definition of derivative, we write this as \(f^{\prime}(1)=\lim _{x \rightarrow 1} \frac{\ln x-\ln 1}{x-1}=1 .\) Explain why this implies that \(\lim _{x \rightarrow 1} \frac{x-1}{\ln x}=1 .\) Finally, substitute \(x=e^{h}.\)
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