/*! 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 7 For Activities 5 through \(16,\)... [FREE SOLUTION] | 91Ó°ÊÓ

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For Activities 5 through \(16,\) evaluate the improper integral. $$ \int_{5}^{\infty} 5\left(0.36^{x}\right) d x $$

Short Answer

Expert verified
The value of the integral is \(-\frac{5 \cdot 0.36^{5}}{\ln(0.36)}\)."

Step by step solution

01

Identify the Integral Type

The given integral \( \int_{5}^{\infty} 5(0.36^{x}) \, dx \) is an improper integral because it has an infinite upper limit.
02

Rewrite as a Limit

To handle the improper integral, rewrite it as a limit: \( \lim_{b \to \infty} \int_{5}^{b} 5(0.36^{x}) \, dx \).
03

Evaluate the Inner Definite Integral

Find the antiderivative of the function. Since the integrand is \( 5 \times 0.36^{x} \), use substitution. The antiderivative is \( \frac{5}{\ln(0.36)} \cdot 0.36^{x} + C \). Evaluate this from \( x = 5 \) to \( x = b \).
04

Apply the Fundamental Theorem of Calculus

Using the antiderivative, calculate: \( \left[ \frac{5}{\ln(0.36)} \cdot 0.36^{x} \right]_{5}^{b} \). This becomes \( \frac{5}{\ln(0.36)} \left( 0.36^{b} - 0.36^{5} \right) \).
05

Evaluate the Limit

Calculate the limit as \( b \to \infty \) of the expression: \( \lim_{b \to \infty} \frac{5}{\ln(0.36)} \left( 0.36^{b} - 0.36^{5} \right) \). Since \( 0.36^{b} \to 0 \) as \( b \to \infty \), the expression simplifies to \( -\frac{5 \cdot 0.36^{5}}{\ln(0.36)} \).
06

Simplify the Expression

Substitute \( 0.36^{5} \) into the expression. The result is the final value of the integral, \( -\frac{5 \cdot 0.36^{5}}{\ln(0.36)} \).

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Key Concepts

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

Antiderivative
An antiderivative of a function is a sort of reverse operation of differentiation. It tells us which function gives us a particular derivative. In simpler terms, if you differentiate an antiderivative, you get the original function back.
To find the antiderivative of a given function, we need to think about which function, when differentiated, will give the function we started with. For instance, consider the function in the original exercise is given by \[ f(x) = 5 \cdot 0.36^x \].
To determine its antiderivative, recall the rule that the antiderivative of a constant raised to a variable power, like 0.36 to x, involves dividing by the natural logarithm of the base. Thus, the antiderivative becomes \[ \frac{5}{\ln(0.36)} \cdot 0.36^x + C \] where C is the constant of integration.
Finding antiderivatives is a crucial step in solving integrals, especially when dealing with improper integrals.
Infinite Limits
Infinite limits are a concept used to deal with integrals that have an infinite bound. These integrals are called improper integrals, and they occur frequently in calculus when we are dealing with situations that extend indefinitely.
In solving an improper integral like the one in our exercise, \[ \int_{5}^{\infty} 5 \cdot 0.36^x \ dx \], we replace the infinite boundary with a variable, say \( b \), and take the limit as \( b \to \infty \).
This approach transforms the improper integral into a limit problem: \[ \lim_{b \to \infty} \int_{5}^{b} 5 \cdot 0.36^x \ dx \].
By using infinite limits, we can assess whether the integral converges (approaches a finite value) or diverges (increases without bound). In our example, since \( 0.36^b \to 0 \) as \( b \to \infty \), the integral converges, simplifying our computation.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a cornerstone of mathematics that links the concepts of differentiation and integration. It essentially states that if you have a continuous function, the derivative of the integral of that function, over some interval, gives you back the function itself.
This theorem has two parts, and in the context of our exercise, we use the second part, which helps to compute a definite integral using antiderivatives.
After finding the antiderivative \( \frac{5}{\ln(0.36)} \cdot 0.36^x + C \), we apply the theorem by evaluating it at the bounds of integration: \[ \left[ \frac{5}{\ln(0.36)} \cdot 0.36^x \right]_{5}^{b} \].
This means substituting the bounds into the antiderivative and subtracting, resulting in \[ \frac{5}{\ln(0.36)} \left( 0.36^b - 0.36^5 \right) \].
The Fundamental Theorem simplifies the process of integration vastly, allowing us to compute definite integrals by working with antiderivatives.

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