/*! 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 48 Use integration by parts to eval... [FREE SOLUTION] | 91Ó°ÊÓ

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Use integration by parts to evaluate the given integral using the following integral formulas where necessary. (You have seen some of these before; all can be checked by differentiating.) $$ \begin{array}{|l|l|} \hline \text { Integral Formula } & \text { Shortcut Version } \\ \hline \int \frac{|x|}{x} d x=|x|+C & \int \frac{|a x+b|}{a x+b} d x=\frac{1}{a}|a x+b|+C \\ \text { Because } \frac{d}{d x}|x|=\frac{|x|}{x} . \\ \int|x| d x=\frac{1}{2} x|x|+C & \int \begin{array}{l} |a x+b| d x \\ =\frac{1}{2 a}(a x+b)|a x+b|+C \end{array} \\ \int x^{2}|x| d x=\frac{1}{4} x^{3}|x|+C & \int \begin{array}{l} (a x+b)^{2}|a x+b| d x \\ =\frac{1}{4 a}(a x+b)^{3}|a x+b|+C \end{array} \\ \hline \int x|x| d x=\frac{1}{3} x^{2}|x|+C & \int \begin{array}{l} (a x \mid b)|a x| b \mid d x \\ =\frac{1}{3 a}(a x+b)^{2}|a x+b|+C \end{array} \\ \hline \end{array} $$ $$\int 3 x \frac{|x+4|}{x+4} d x$$

Short Answer

Expert verified
The short answer to the question is: \(\frac{1}{3}(3x+4)^2|3x+4|+C\).

Step by step solution

01

Identify the integral formula

For the given integral, we can identify that it falls under the format $$\int(ax+b)\frac{|ax+b|}{ax+b}dx$$ from our table of integral formulas. The corresponding shortcut is: $$\int(ax+b)\frac{|ax+b|}{ax+b}dx=\frac{1}{a}(ax+b)^2|ax+b|+C$$, where $$a$$ and $$b$$ are constants. In our problem, we have $$a = 3$$ and $$b = 4$$.
02

Apply the corresponding shortcut formula

Using the identified shortcut formula and our values for $$a$$ and $$b$$, we can directly evaluate the integral as follows: $$\int 3x \frac{|x+4|}{x+4}dx = \frac{1}{3}(3x+4)^2|3x+4|+C$$. Hence, the solution of the integral is: $$\frac{1}{3}(3x+4)^2|3x+4|+C$$.

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Key Concepts

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

Integral Formulas
Integral formulas provide a convenient way to find the antiderivative of certain functions without having to perform complex integrations manually. They serve as shortcuts, allowing us to quickly solve integrals by identifying their pattern and matching it to a known formula. This is similar to how algebraic identities simplify complex expressions.

When dealing with absolute values in integrals, it becomes essential to use specific integral formulas. With these formulas, you can
  • Avoid tedious calculations.
  • Ensure accuracy in integration, especially when absolute values are involved.
  • Recognize patterns more swiftly, which speeds up problem-solving.
In the case where the integral involves expressions like \( \int (ax+b)\frac{|ax+b|}{ax+b}dx \), you can use ready-made formulas such as \( \frac{1}{a}(ax+b)^2|ax+b|+C \) to find the integral directly.

These formulas save time and effort and are indispensable in dealing with more complex integrals swiftly.
Absolute Value Integration
Absolute value integration involves dealing with the function \(|x|\), which isn't directly differentiable at \(x=0\). This is because \(|x|\) changes direction at zero, making integration slightly more complex than with standard functions.

To integrate a function with an absolute value, we often break it into two cases — one for when the expression inside the absolute value is positive and one for when it is negative. However, using integral formulas designed specifically for absolute values can bypass this needlessly tedious step.

In cases like \(\int\frac{|x+4|}{x+4}dx\), it's crucial to recognize that these absolute value expressions simplify the integration process. The formulas handle the discontinuity at \(x=-4\), allowing integration without splitting the interval.

By knowing specific shortcuts for absolute value integrals, you can avoid lengthy manual calculations, letting the formulas do the heavy lifting.
Integration Techniques
Integration techniques such as integration by parts, substitution, and the use of integral formulas are essential strategies in calculus. They help tackle integrals that seem too complex or labor-intensive for basic methods.

Integration by parts is particularly useful when dealing with products of functions. It is based on the product rule for differentiation and provides a way to transform a complex integral into simpler parts. This method is formalized by the formula:
  • \(\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx\).
  • Here, \(u\) and \(v\) are functions of \(x\), carefully chosen to simplify the integration process.
However, in some cases, readily available integral formulas, like the ones for absolute values, can further ease the solution. This cuts down on the work required and provides us with a direct path to finding the integral.

Mastering these techniques and knowing when to apply each one enables efficient problem-solving, making even the most challenging integrals accessible.

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Most popular questions from this chapter

Electric Circuits The flow of current \(i(t)\) in an electric circuit without capacitance satisfies the linear differential equation $$ L \frac{d i}{d t}+R i=V(t) $$ where \(L\) and \(R\) are constants (the inductance and resistance respectively) and \(V(t)\) is the applied voltage. (See figure.) If the voltage is supplied by a 10 -volt battery and the switch is turned on at time \(t=1\), then the voltage \(V\) is a step function that jumps from 0 to 10 at \(t=1: V(t)=5\left[1+\frac{|t-1|}{t-1}\right]\). Find the current as a function of time for \(L=R=1 .\) Use a grapher to plot the resulting current as a function of time. (Assume there is no current flowing at time \(t=0 .\) [Use the following integral formula: \(\int\left[1+\frac{|t-1|}{t-1}\right] e^{t} d t=\) \(\left.\left[1+\frac{|t-1|}{t-1}\right]\left(e^{t}-e\right)+C .\right]\)

Laplace Transforms The Laplace Transform \(F(x)\) of a function \(f(t)\) is given by the formula $$ F(x)=\int_{0}^{+\infty} f(t) e^{-x t} d t \quad(x>0) $$ a. Find \(F(x)\) for \(f(t)=1\) and for \(f(t)=t\). b. Find a formula for \(F(x)\) if \(f(t)=t^{n}(n=1,2,3, \ldots)\). c. Find a formula for \(F(x)\) if \(f(t)=e^{a t}(a\) constant).

Linear Differential Equationsare based on first order linear differential equations with constant coefficients. These have the form $$ \frac{d y}{d t}+p y=f(t) \quad(p \text { constant }) $$ and the general solution is $$ y=e^{-p t} \int f(t) e^{p t} d t . \quad \text { (Check this by substituting!) } $$ Solve the linear differential equation $$ 2 \frac{d y}{d t}-y=2 t ; y=1 \text { when } t=0 . $$

Newton's Law of Cooling For coffee in a paper cup, suppose \(k \approx 0.08\) with time measured in minutes. (a) Use Newton's Law of Cooling to predict the temperature of the coffee, initially at a temperature of \(210^{\circ} \mathrm{F}\), that is left to sit in a room at \(60^{\circ} \mathrm{F}\). (b) When will the coffee have cooled to \(70^{\circ} \mathrm{F}\) ? HINI [See Example 4.]

Valuing Future Income Max was injured and can no longer work. As a result of a lawsuit, he is to be awarded the present value of the income he would have received over the next 30 years. His income at the time he was injured was $$\$ 30,000$$ per year, increasing by $$\$ 1,500$$ per year. What will be the amount of his award, assuming continuous income and a \(6 \%\) interest rate?

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