/*! 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 90 Evaluate the integrals. \begin... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 90

Evaluate the integrals. \begin{equation}\int_{-2}^{0} 5^{-\theta} d \theta\end{equation}

Short Answer

Expert verified
The integral evaluates to \( \frac{24}{\ln(5)} \).

Step by step solution

01

Identify the function inside the integral

The given integral is \( \int_{-2}^{0} 5^{-\theta} \, d\theta \). This means we need to integrate the function \( 5^{-\theta} \) with respect to \( \theta \).
02

Determine the antiderivative

To integrate \( 5^{-\theta} \), we need to use the formula for integrating exponential functions. The antiderivative of \( a^{-\theta} \) with respect to \( \theta \) is \( \frac{a^{-\theta}}{-\ln(a)} + C \), as long as \( a > 0 \) and \( a eq 1 \). Here, \( a = 5 \), so the antiderivative is \( \frac{5^{-\theta}}{-\ln(5)} \).
03

Apply the limits of integration

We now evaluate the definite integral using the antiderivative: \[ \left[ \frac{5^{-\theta}}{-\ln(5)} \right]_{-2}^{0} = \frac{5^{0}}{-\ln(5)} - \frac{5^{-(-2)}}{-\ln(5)} \].
04

Simplify the expression

Calculate \( 5^{0} \) which is 1, and \( 5^{2} \) which is 25. Substitute these into the expression from Step 3: \[ \frac{1}{-\ln(5)} - \frac{25}{-\ln(5)} = \frac{1 - 25}{-\ln(5)} = \frac{-24}{-\ln(5)} = \frac{24}{\ln(5)} \]. This is the value of the integral.

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Key Concepts

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

Exponential Functions
Exponential functions are fundamental in mathematics, commonly denoted as \( a^x \), where \( a \) is a positive constant base and \( x \) is the exponent. They have distinct properties that make them incredibly useful, such as rapid growth or decay depending on the sign of the exponent.

Some key characteristics include:
  • If \( a > 1 \), the function increases rapidly as \( x \) increases.
  • If \( 0 < a < 1 \), the function decreases quickly as \( x \) increases.
  • The graph of an exponential function has a horizontal asymptote typically at \( y=0 \).
In the context of integration, especially for negative exponents like \( 5^{-\theta} \), it involves understanding how these functions change over an interval. The base \( a \), here being 5, dictates the behavior of the function over specified bounds in the integral, such as from \( -2 \) to \( 0 \).
Antiderivative
The antiderivative, or the indefinite integral, is the reverse process of differentiation. For a function \( f(x) \), the antiderivative is a function \( F(x) \) such that \( F'(x) = f(x) \). It's noted by the integral symbol with no specified limits: \( \int f(x) \, dx \).

Calculating the antiderivative:
  • For exponential functions like \( a^{-\theta} \), utilize the formula \( \frac{a^{-\theta}}{-\ln(a)} + C \).
  • \( C \) is the constant of integration, symbolizing that indefinite integrals can have multiple solutions.
In our exercise, to find the antiderivative of \( 5^{-\theta} \), we applied this formula to find \( \frac{5^{-\theta}}{-\ln(5)} \), highlighting the necessity to understand and apply the formula correctly for definite integrals.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus bridges the concept of differentiation with integration, providing a method for evaluating definite integrals. It states that if \( F \) is an antiderivative of \( f \) over an interval \([a, b]\), then: \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \].

Application steps include:
  • Find the antiderivative \( F(x) \) of \( f(x) \).
  • Evaluate \( F(x) \) at both bounds \( a \) and \( b \).
  • Subtract the two results: \( F(b) - F(a) \).
In the given problem, after finding the antiderivative \( \frac{5^{-\theta}}{-\ln(5)} \), we applied this theorem. We calculated the difference between its values at the boundary points, \( 0 \) and \( -2 \), to find the definite integral: \[ \frac{24}{\ln(5)} \]. This demonstrates the power of this theorem in simplifying calculations and providing exact answers.

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Most popular questions from this chapter

For problems \(43-46\) use implicit differentiation to find \(\frac{d y}{d x}\) at the given point \(P .\) $$ 16\left(\tan ^{-1} 3 y\right)^{2}+9\left(\tan ^{-1} 2 x\right)^{2}=2 \pi^{2} ; \quad P\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right) $$

Find the area of the region between the curve \(y=2 x /\left(1+x^{2}\right)\) and the interval \(-2 \leq x \leq 2\) of the \(x\) -axis.

Evaluate the integrals in Exercises \(71-84\) $$ \int_{1}^{2} \frac{8 d x}{x^{2}-2 x+2} $$

In Exercises \(21-42,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\sec ^{-1} 5 s $$

In Exercises \(21-42,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\csc ^{-1}\left(x^{2}+1\right), x>0 $$
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