Problem 35
Question
Evaluate the integrals. \begin{equation}\int_{\ln 2}^{\ln 3} e^{x} d x\end{equation}
Step-by-Step Solution
Verified Answer
The value of the integral is 1.
1Step 1: Understanding the Problem
We need to evaluate the definite integral \[\int_{\ln 2}^{\ln 3} e^{x} dx\]This involves finding the area under the curve of the function \(e^x\) from \(x = \ln 2\) to \(x = \ln 3\).
2Step 2: Finding the Antiderivative
Identify the antiderivative of \(e^x\). Since the derivative of \(e^x\) is itself \(e^x\), the antiderivative of \(e^x\) is also \(e^x\).
3Step 3: Evaluating the Definite Integral
Use the Fundamental Theorem of Calculus to evaluate the integral:\[F(b) - F(a)\] where \(F(x) = e^x\), \(a = \ln 2\) and \(b = \ln 3\).
4Step 4: Substituting the Upper Limit
Substitute the upper limit into the antiderivative:\[F(\ln 3) = e^{\ln 3} = 3\]
5Step 5: Substituting the Lower Limit
Substitute the lower limit into the antiderivative:\[F(\ln 2) = e^{\ln 2} = 2\]
6Step 6: Calculating the Difference
Compute the difference between the results of the upper limit and lower limit:\[F(\ln 3) - F(\ln 2) = 3 - 2 = 1\].
Key Concepts
AntiderivativeFundamental Theorem of CalculusExponential Function
Antiderivative
An antiderivative is essentially the reverse process of differentiation. In the context of our exercise, we are dealing with the function \( e^x \), which is unique because its derivative is the same as the function itself.
Therefore, the antiderivative of \( e^x \) is \( e^x + C \), where \( C \) is an arbitrary constant.
Antiderivatives play a crucial role when evaluating definite integrals as they are used to help find the area under a curve between two points.
Therefore, the antiderivative of \( e^x \) is \( e^x + C \), where \( C \) is an arbitrary constant.
Antiderivatives play a crucial role when evaluating definite integrals as they are used to help find the area under a curve between two points.
- Think of the antiderivative as a way to "undo" the derivative.
- Note that different functions have different antiderivatives, but \( e^x \) is special for its simplicity.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links the processes of differentiation and integration. It's split into two parts, but the one we focus on here allows us to evaluate definite integrals efficiently.
This theorem states that if \( F \) is the antiderivative of \( f \) over an interval, then:\[\int_a^b f(x) dx = F(b) - F(a)\]This means that to find the definite integral from \( a \) to \( b \), you compute the antiderivative at \( b \) and subtract the antiderivative at \( a \).
This theorem states that if \( F \) is the antiderivative of \( f \) over an interval, then:\[\int_a^b f(x) dx = F(b) - F(a)\]This means that to find the definite integral from \( a \) to \( b \), you compute the antiderivative at \( b \) and subtract the antiderivative at \( a \).
- This simplifies the complex process of finding areas under curves.
- It's one of the cornerstone principles of calculus that sets the foundation for justifying why derivatives and integrals are inverses.
Exponential Function
Exponential functions are striking for their specific properties and common applications. The function \( e^x \) is the most well-known exponential function.
Its base "e" is approximately equal to 2.71828 and arises naturally in various mathematical contexts, from growth processes to physics.
Understanding \( e^x \)'s properties helps you grasp why its antiderivative remains \( e^x \), simplifying our definite integral calculation in the given exercise, making our solution direct and clear.
Its base "e" is approximately equal to 2.71828 and arises naturally in various mathematical contexts, from growth processes to physics.
- The function \( e^x \) grows faster than polynomial functions as \( x \) increases.
- Across the real number line, it is always positive, never zero.
Understanding \( e^x \)'s properties helps you grasp why its antiderivative remains \( e^x \), simplifying our definite integral calculation in the given exercise, making our solution direct and clear.
Other exercises in this chapter
Problem 35
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