Question
Evaluate the following definite integral using the Fundamental Theorem of Calculus. \( \frac{1}{10} \int_{0}^{\ln 2} e^{x} d x \) \( \frac{1}{10} \int_{0}^{\ln 2} e^{x} d x=\square \) (Type an integer or a simplified fraction.)
Ask by Ortiz Schneider. in the United States
Jan 23,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
\(\frac{1}{10}\)
Solution
To evaluate the definite integral \( \frac{1}{10} \int_{0}^{\ln 2} e^{x} \, dx \) using the Fundamental Theorem of Calculus, follow these steps:
1. **Find the Antiderivative:**
The antiderivative of \( e^x \) is \( e^x \).
2. **Evaluate the Definite Integral:**
\[
\int_{0}^{\ln 2} e^{x} \, dx = e^{\ln 2} - e^{0} = 2 - 1 = 1
\]
3. **Multiply by the Constant:**
\[
\frac{1}{10} \times 1 = \frac{1}{10}
\]
**Final Answer:**
\(\frac{1}{10}\)
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Simplify this solution Beyond the Answer
To evaluate the integral \( \frac{1}{10} \int_{0}^{\ln 2} e^{x} d x \), we first need to solve the integral \( \int e^{x} d x \), which is simply \( e^{x} + C \). Now, applying the Fundamental Theorem of Calculus: \[ \int_{0}^{\ln 2} e^{x} d x = \left[ e^{x} \right]_{0}^{\ln 2} = e^{\ln 2} - e^{0} = 2 - 1 = 1. \] Now, we multiply this result by \( \frac{1}{10} \): \[ \frac{1}{10} \int_{0}^{\ln 2} e^{x} d x = \frac{1}{10} \cdot 1 = \frac{1}{10}. \] So the final answer is \[ \frac{1}{10}. \] Hence \( \frac{1}{10} \int_{0}^{\ln 2} e^{x} d x=\frac{1}{10} \).
