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Chapter 7: Problem 47

Evaluate the integrals. \(\int 5^{x} d x\)

Short Answer

Expert verified
The integral is \(\int 5^x \, dx = \frac{5^x}{\ln 5} + C\).

Step by step solution

01

Identify the Type of Integral

This integral involves an exponential function, specifically in the form of base 5 raised to a variable exponent. Therefore, we will use the formula for integrating exponential functions of the form \(\int a^{x} \, dx = \frac{a^{x}}{\ln a} + C\).
02

Apply the Exponential Integral Formula

Using the formula from Step 1, substitute \(a = 5\) into \(\int 5^{x} \, dx = \frac{5^{x}}{\ln 5} + C\), where \(C\) is the constant of integration.

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Key Concepts

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

Exponential Function
An exponential function is a powerful mathematical concept that occurs throughout various mathematical and real-world applications. The function typically takes the form \(a^x\), where \(a\) is a constant base and \(x\) is the exponent. In this type of function:
  • The base \(a\) is a positive constant not equal to 1.
  • The function grows or decays rapidly due to the power of the exponent \(x\).
  • If \(a > 1\), the function exhibits exponential growth.
  • If \(0 < a < 1\), the function shows exponential decay.
Exponential functions are key in fields such as biology for modeling population growth, finance for compound interest calculations, and physics for radioactive decay analysis. Integrating exponential functions requires a specific approach since their derivatives are proportional to the function itself.
Integration Techniques
Integration techniques are mathematical strategies used to solve integrals, which are operations to find the area under curves represented by functions. When dealing with exponential functions, like \(5^x\), a specific technique for integration is employed. This involves using a known formula:
  • For exponential functions in the form of \(a^x\), the integral is \(\int a^x \, dx = \frac{a^x}{\ln a} + C\).
  • The constant \(a\) gets raised to the power of \(x\), with the result divided by the natural logarithm of \(a\).
  • This formula takes into account the unique properties of exponential growth and decay.
Integration can seem complex, but by applying these strategies step-by-step, you can navigate through the problem to find a solution effectively. This technique is one of many that simplifies the process of finding integrals, especially in calculus problems involving growth and decay.
Constant of Integration
The constant of integration, denoted as \(C\), plays a crucial role in indefinite integrals. While definite integrals provide a specific value, indefinite integrals yield a broad family of functions.
  • Every integral requires adding a constant of integration because the process of differentiation eliminates constants; we must account for any constant that could have existed.
  • This constant represents the idea that there could be several functions that differentiate to the same derivative.
  • If you take the derivative of any of these functions, the result will be the same, but their integral representations slightly differ by a constant value.
Remember, the constant \(C\) becomes essential in equations and modeling because it accounts for initial conditions or offsets, ensuring the solved integral accurately reflects the starting scenario or benchmark in specific problems. Thus, it completes the integration process by encapsulating all possible original functions.

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Most popular questions from this chapter

Evaluate the integrals in Exercises \(71-94\) $$ \int_{0}^{\ln \sqrt{3}} \frac{e^{x} d x}{1+e^{2 x}} $$

Evaluate the integrals in Exercises \(71-94\) $$ \int \frac{d x}{1+(3 x+1)^{2}} $$

Evaluate the integrals in Exercises \(105-112\) . $$ \int \frac{\left(\sin ^{-1} x\right)^{2} d x}{\sqrt{1-x^{2}}} $$

When hyperbolic function keys are not available on a calculator, it is still possible to evaluate the inverse hyperbolic functions by expressing them as logarithms, as shown here. $$ \begin{array}{rlrl}{\sinh ^{-1} x} & {=\ln \left(x+\sqrt{x^{2}+1}\right),} & {-\infty < x < \infty} \\ {\cosh ^{-1} x} & {=\ln \left(x+\sqrt{x^{2}+1}\right),} & {x \geq 1} \\ {\tanh ^{-1} x} & {=\frac{1}{2} \ln \frac{1+x}{1-x},} & {|x| < 1} \\ {\operatorname{sech}^{-1} x} & {=\ln \left(\frac{1+\sqrt{1-x^{2}}}{x}\right),} & {0 < x \leq 1} \\\ {\operatorname{csch}^{-1} x} & {=\ln \left(\frac{1}{x}+\frac{\sqrt{1+x^{2}}}{|x|}\right),} & {x \neq 0} \\\ {\operatorname{coth}^{-1} x} & {=\frac{1}{2} \ln \frac{x+1}{x-1},} & {|x| > 1}\end{array} $$ Use the formulas given above to express the numbers in Exercises \(61-66\) in terms of natural logarithms. $$ \cosh ^{-1}(5 / 3) $$

Tractor trailers and the tractrix When a tractor trailer turns into a cross street or driveway, its rear wheels follow a curve like the one shown here. (This is why the rear wheels sometimes ride up over the curb.) We can find an equation for the curve if we picture the rear wheels as a mass \(M\) at the point \((1,0)\) on the \(x\) -axis attached by a rod of unit length to a point \(P\) representing the cab at the origin. As the point \(P\) moves up the \(y\) -axis, it drags \(M\) along behind it. The curve traced by \(M-\) called a tractrix from the Latin word tractum, for "drag" - can be shown to be the graph of the function \(y=f(x)\) that solves the initial value problem $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d y}{d x}=-\frac{1}{x \sqrt{1-x^{2}}}+\frac{x}{\sqrt{1-x^{2}}}} \\ {\text { Initial condition: }} & {y=0 \quad \text { when } \quad x=1}\end{array} $$ Solve the initial value problem to find an equation for the curve. (You need an inverse hyperbolic function.) Graph cannot copy
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