/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 Find the indefinite integral. ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the indefinite integral. $$\int\left(6 x^{3}+\frac{3}{x^{2}}-x\right) d x$$

Short Answer

Expert verified
The indefinite integral of the given function is: \( \int\left(6 x^{3}+\frac{3}{x^{2}}-x\right) d x = \frac{3}{2}x^4 - \frac{3}{x} - \frac{1}{2}x^2 + C \)

Step by step solution

01

Break the function into individual terms

We are given the function to integrate: \( \int\left(6 x^{3}+\frac{3}{x^{2}}-x\right) d x \) To find the integral, we need to break this function into individual terms: \( \int{6x^3dx} + \int{\frac{3}{x^2}dx} - \int{xdx} \)
02

Integrate each term

Now, we integrate each term separately: For the first term: \( \int{6x^3dx} \) Use the power rule: \(\int{x^n dx} = \frac{x^{n+1}}{n+1} + C\) So, \( \int{6x^3dx} = 6\frac{x^{3+1}}{3+1} + C_1 = \frac{3}{2}x^4 + C_1 \) For the second term: \( \int{\frac{3}{x^2}dx} \) Rewrite the function as: \( \int{3x^{-2}dx} \) Use the power rule again: \( \int{3x^{-2}dx} = 3\frac{x^{-2+1}}{-2+1} + C_2 = -3x^{-1} + C_2 = -\frac{3}{x} + C_2 \) For the third term: \( \int{xdx} \) Using the power rule: \( \int{xdx} = \frac{x^{1+1}}{1+1} + C_3 = \frac{1}{2}x^2 + C_3 \)
03

Combine the integrated terms

Now that we have integrated each term, we need to combine them to find the indefinite integral of the given function: \( \int\left(6 x^{3}+\frac{3}{x^{2}}-x\right) d x = \frac{3}{2}x^4 - \frac{3}{x} - \frac{1}{2}x^2 + C \) where C is the constant of integration, equal to the sum of constants, i.e., C = C_1 + C_2 + C_3. So the indefinite integral of the given function is: \( \int\left(6 x^{3}+\frac{3}{x^{2}}-x\right) d x = \frac{3}{2}x^4 - \frac{3}{x} - \frac{1}{2}x^2 + C \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Indefinite Integral
Integrals are fundamental in calculus and help us find the area under curves and accumulated quantities. An *indefinite integral* is essentially the opposite of differentiation. It gives us a function whose derivative is the original function we started with. This function is called the *antiderivative*. An indefinite integral includes a constant of integration, often denoted as 'C', because the process of integration only finds one antiderivative — but there are infinitely many since constants vanish when differentiated.

When you see a problem like \( \int (6x^3 + \frac{3}{x^2} - x) dx \), the task is to determine the antiderivative for each term within the integral. The integral's symbol is an elongated 'S', standing for 'summation': it's used to denote the process of integration, where we are summing up an infinite number of infinitely small pieces.
Power Rule in Integration
The *Power Rule in Integration* is analogous to the Power Rule in Derivatives and is a very handy tool for finding antiderivatives of polynomial functions. The rule is straightforward: for any power of \( x \), denoted as \( x^n \), the integral is given by:
\[ \int x^n dx = \frac{x^{n+1}}{n+1} + C \]
This formula works as long as \( n eq -1 \), because when \( n = -1 \), we have a special case where the integral of \( x^{-1} \) is \( \ln|x| \). The Power Rule allows us to efficiently integrate each term of a polynomial separately. For example, to integrate \( 6x^3 \), apply the Power Rule:
  • Increase the exponent by 1, turning \( x^3 \) into \( x^4 \).
  • Divide by the new exponent, resulting in \( \frac{6}{4}x^4 \) which simplifies to \( \frac{3}{2}x^4 \).
Remember, each term in a function can be integrated separately using this rule, which simplifies calculations greatly.
Integration of Polynomial Functions
*Polynomial functions* consist of multiple terms where each term is a constant multiplied by a variable raised to a non-negative integer power. Integrating polynomial functions involves applying the Power Rule to each individual term. For example, \( 6x^3 \) and \( -x \) in the function \( 6x^3 + \frac{3}{x^2} - x \) are already written with their powers, but \( \frac{3}{x^2} \) needs to be rewritten as \( 3x^{-2} \) for easier integration.

Thus after rewriting, you integrate each term separately:
  • For \( 6x^3 \), you get \( \frac{3}{2}x^4 \).
  • For \( 3x^{-2} \), the integral is \( -3x^{-1} \) or equivalently \( -\frac{3}{x} \).
  • The term \( -x \) integrates to \( -\frac{1}{2}x^2 \).
Each of these calculations follows directly from applying the Power Rule.
Total integration becomes combining these integrated terms together, plus the constant of integration \( C \). The presence of \( C \) accounts for any constant that could be added to each antiderivative, signifying the continuous family of possible solutions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In an endeavor to curb population growth in a Southeast Asian island state, the government has decided to launch an extensive propaganda campaign. Without curbs, the government expects the rate of population growth to have been $$ 60 e^{0.02 t} $$ thousand people/year, \(t\) yr from now, over the next 5 yr. However, successful implementation of the proposed campaign is expected to result in a population growth rate of $$ -t^{2}+60 $$ thousand people/year, \(t\) yr from now, over the next 5 yr. Assuming that the campaign is mounted, how many fewer people will there be in that country 5 yr from now than there would have been if no curbs had been imposed?

The demand function for a certain brand of \(\mathrm{CD}\) is given by $$ p=-0.01 x^{2}-0.2 x+8 $$ where \(p\) is the wholesale unit price in dollars and \(x\) is the quantity demanded each week, measured in units of a thousand. Determine the consumers' surplus if the wholesale market price is set at \(\$ 5 /\) disc.

The increase in carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) in the atmosphere is a major cause of global warming. Using data obtained by Charles David Keeling, professor at Scripps Institution of Oceanography, the average amount of \(\mathrm{CO}_{2}\) in the atmosphere from 1958 through 2007 is approximated by \(A(t)=0.010716 t^{2}+0.8212 t+313.4 \quad(1 \leq t \leq 50)\) where \(A(t)\) is measured in parts per million volume (ppmv) and \(t\) in years, with \(t=1\) corresponding to 1958 . Find the average rate of increase of the average amount of \(\mathrm{CO}_{2}\) in the atmosphere from 1958 through 2007 .

Show that the area of a region \(R\) bounded above by the graph of a function \(f\) and below by the graph of a function \(g\) from \(x=a\) to \(x=b\) is given by $$ \int_{a}^{b}[f(x)-g(x)] d x $$ Hint: The validity of the formula was verified earlier for the case when both \(f\) and \(g\) were nonnegative. Now, let \(f\) and \(g\) be two functions such that \(f(x) \geq g(x)\) for \(a \leq x \leq b\). Then, there exists some nonnegative constant \(c\) such that the curves \(y=f(x)+c\) and \(\bar{y}=\) \(g(x)+c\) are translated in the \(y\) -direction in such a way that the region \(R^{r}\) has the same area as the region \(R\) (see the accompanying figures). Show that the area of \(R^{\prime}\) is given by $$ \int_{a}^{b}\\{[f(x)+c]-[g(x)+c]\\} d x=\int_{a}^{b}[f(x)-g(x)] d x $$

Sketch the graph and find the area of the region bounded below by the graph of each function and above by the \(x\) -axis from \(x=a\) to \(x=b\). $$f(x)=\frac{1}{2} x-\sqrt{x} ; a=0, b=4$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.