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Chapter 6: Problem 75

State (without carrying them out) two different methods to find \(\int \ln t d t\).

Short Answer

Expert verified
Use integration by parts or substitution with change of variables.

Step by step solution

01

Integration by Parts Formula

To solve the integral \( \int \ln t \, dt \), you can use the integration by parts formula, which is given by:\[\int u \, dv = uv - \int v \, du\]In this formula, choose \( u = \ln t \) and \( dv = dt \). Then find \( du = \frac{1}{t} \, dt \) and \( v = t \).
02

Substitution Method

Another method is via substitution. Recognize the integral:\[\int \ln t \, dt\]You can substitute \( t = e^u \), thus \( dt = e^u \, du \) and \( \ln t = u \). This transforms the integral into \( \int u e^u \, du \), which can then be solved using integration by parts.

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Key Concepts

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

Integration by Parts
Integration by Parts is like the reverse of the product rule for derivatives. It helps us integrate products of functions by transforming them into simpler integrals.

The formula is given by:
  • \[\int u \, dv = uv - \int v \, du\]
Here's how it works:
  • Choose the function in your integral as "\( u \)", which should simplify upon differentiation. For \( \int \ln t \, dt\), \( u = \ln t \).
  • Let the remaining part of the integral be "\( dv \)", so \( dv = dt \).
  • Find \( du \) by differentiating \( u \), which gives \( du = \frac{1}{t} \, dt \).
  • Integrate \( dv \) to get \( v \), which results in \( v = t \).
  • Substitute into the formula, yielding \( t \ln t - \int t \frac{1}{t} \, dt \), which simplifies the integral.
This method is useful, especially when functions are difficult to integrate directly, like logarithmic functions.
Substitution Method
The substitution method, or "u-substitution," simplifies integrals by changing variables. It's particularly helpful when faced with compositions of functions.

For the integral \( \int \ln t \, dt \), consider:
  • Set \( t = e^u \). It's a clever substitution since it simplifies \( \ln t \) to \( u \).
  • Differentiate \( t = e^u \) to find \( dt = e^u \, du \).
  • After substitution, the integral becomes \( \int u \, e^u \, du \).
  • This new integral is often easier to handle and can be solved using integration techniques such as integration by parts.
This method transforms the integral into a form that can be straightforwardly evaluated, especially when dealing with exponential or trigonometric functions.
Logarithmic Functions
Logarithmic functions, often noted as \( \ln(x) \), are the inverse of exponential functions. They play a significant role in calculus, particularly in integration techniques.

Understanding them helps when choosing integration strategies:
  • Logs transform products into sums: \( \ln(ab) = \ln a + \ln b \).
  • They simplify powers: \( \ln(a^b) = b \ln a \).
  • When integrating \( \ln t \), it doesn't fit basic formulas, necessitating techniques like integration by parts.
These properties make logarithms tricky but rewarding to work with. They often appear in problems involving growth and decay, making them essential in both mathematics and real-world applications.

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Most popular questions from this chapter

Find the solution \(y(t)\) by recognizing each differential equation as determining unlimited, limited, or logistic growth, and then finding the constants. \(\begin{aligned} y^{\prime} &=y(2-y) \\ y(0) &=4 \end{aligned}\)

Since preferred stock can remain outstanding indefinitely, the present value per share is the limit of the present value of an annuity* paying that share's dividend \(D\) at interest rate \(r\) : $$ \left(\begin{array}{c} \text { Present } \\ \text { value } \end{array}\right)=\lim _{t \rightarrow \infty} D\left(\frac{1-(1+r)^{-t}}{r}\right) $$ Find this limit in terms of \(D\) and \(r\).

Find the solution \(y(t)\) by recognizing each differential equation as determining unlimited, limited, or logistic growth, and then finding the constants. \(\begin{aligned} y^{\prime} &=\frac{2}{3}(1-y) \\ y(0) &=0 \end{aligned}\)

Find the solution \(y(t)\) by recognizing each differential equation as determining unlimited, limited, or logistic growth, and then finding the constants. \(y^{\prime}=y(1-y)\) \(y(0)=\frac{1}{2}\)

True or False: If \(\int_{-\infty}^{x} f(x) d x\) converges, then \(\int_{0}^{x} f(x) d x\) converges.
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