Problem 27 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int_{-1}^{1} 10^{x} d x$$

Step-by-Step Solution

Answer

Answer: The value of the definite integral is $$\frac{99}{10\ln 10}$$.

Step 1: Find the indefinite integral of the function

To find the indefinite integral of the function $$10^x$$ with respect to $$x$$, we can use the formula $$\int e^{u} d u=e^{u}$$ for exponential functions but first, we need to rewrite $$10^x$$ in terms of $$e$$ which can be done by using the formula $$a^x = e^{(\ln a)x}$$, so we have: $$10^x = e^{(\ln 10)x}$$ Now, let $$u = (\ln 10)x$$. We can find the integral by doing a substitution: $$\int 10^x dx = \int e^u \frac{1}{\ln 10} du = \frac{1}{\ln 10}\int e^u du$$ Applying the integral formula for exponential functions: $$\frac{1}{\ln 10} e^u$$ Now, we need to substitute $$u$$ back to get the indefinite integral in terms of $$x$$: $$\frac{1}{\ln 10} e^{(\ln 10)x} = \frac{1}{\ln 10} 10^x$$ So, the indefinite integral is $$\frac{1}{\ln 10} 10^x + C$$, where $$C$$ is the constant of integration.

Step 2: Apply the limits of integration

Now, we need to apply the limits of integration $$[-1, 1]$$ to find the definite integral: $$\int_{-1}^1 10^x dx = \left[\frac{1}{\ln 10} 10^x \right]_{-1}^1$$ To calculate this expression, we will substitute the limits of integration: $$\frac{1}{\ln 10} \left[10^1 - 10^{-1} \right] = \frac{1}{\ln 10} \left[10 - \frac{1}{10}\right] = \frac{1}{\ln 10} \left[\frac{99}{10}\right]$$ So, the definite integral is: $$\int_{-1}^1 10^x dx = \frac{99}{10\ln 10}$$

Key Concepts

Indefinite Integral

An indefinite integral, often referred to as an antiderivative, is the reverse process of differentiation. It is a function that describes a family of functions which, when differentiated, yield the original function. When evaluating an indefinite integral, the result typically includes a constant of integration, represented by \( C \).

• The indefinite integral is written as \( \int f(x) \, dx \), where \( f(x) \) is the function you are integrating.
• The process is akin to finding the original function from a derivative, meaning that it "undoes" the derivative.

In our example, we looked for the indefinite integral of \( 10^x \). Through substitution, we first expressed \( 10^x \) in terms of the natural exponential function, making it more manageable to integrate. The indefinite integral was calculated as \( \frac{1}{\ln 10} \cdot 10^x + C \), demonstrating how transformation and substitution help solve such integrals easily.

Exponential Functions

Exponential functions are an important class of functions characterized by a constant raised to the power of a variable. The most commonly used base is \( e \), the natural logarithm base, but other bases such as 10 or 2 are also used frequently.

• An exponential function can be expressed in the form \( a^x \), where \( a \) is the base and \( x \) is the exponent.
• Exponential functions grow rapidly and are used widely in contexts like population growth, radioactive decay, and financial modeling.

In our exercise, we dealt with \( 10^x \), an exponential function with a base of 10. To integrate it, we converted it into an equivalent expression with the base \( e \) using the identity \( a^x = e^{(\ln a)x} \). This allows us to apply standard methods of integration for exponential functions effectively.

Substitution Method

The substitution method is a technique used to simplify integration, especially useful when dealing with composite functions or when an integral appears complicated. It involves changing variables to simplify the integral.

• This method relies on the formula \( \int f(g(x))g'(x) \, dx = \int f(u) \, du \), where \( u = g(x) \) and \( du = g'(x) \, dx \).
• Through substitution, the integral becomes easier to handle, and you revert back to the original variable once integration is complete.

For our problem, we used substitution by setting \( u = (\ln 10)x \). This allowed us to rewrite the integral of \( e^u \) as \( \frac{1}{\ln 10} \int e^u \, du \). Once we integrated \( e^u \), we substituted back to express our answer in terms of \( x \), achieving a result that aligns with the original problem context.