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Chapter 5: Problem 86

Evaluate the following definite integrals using the Fundamental Theorem of Calculus. $$\frac{1}{2} \int_{0}^{\ln 2} e^{x} d x$$

Short Answer

Expert verified
Question: Evaluate the definite integral $$\frac{1}{2}\int_{0}^{\ln 2} e^{x} dx$$ using the Fundamental Theorem of Calculus. Answer: The definite integral $$\frac{1}{2}\int_{0}^{\ln 2} e^{x} dx$$ evaluates to $$\frac{1}{2}$$.

Step by step solution

01

Identify the function and the interval

First, let's identify the function we need to integrate and the interval over which we're integrating. We have the function $$e^x$$ and the interval $$[0, \ln 2]$$. The given integral is: $$\frac{1}{2}\int_{0}^{\ln 2} e^{x} dx$$
02

Find the antiderivative

Next, we need to find an antiderivative of $$e^x$$. The antiderivative of $$e^x$$ is simply $$e^x$$, because the derivative of $$e^x$$ is also $$e^x$$. So, we have: $$\frac{1}{2} [e^x]$$
03

Evaluate the antiderivative at the endpoints

Now we need to evaluate the antiderivative at the endpoints of the interval: $$\ln 2$$ and $$0$$. To do this, we will subtract the antiderivative evaluated at $$0$$ from the antiderivative evaluated at $$\ln 2$$: $$\frac{1}{2} [e^{\ln 2} - e^{0}]$$
04

Simplify the expression

To simplify our expression further, we can use the fact that $$e^{\ln 2} = 2$$ and $$e^{0} = 1$$: $$\frac{1}{2} [2 - 1]$$
05

Compute the final result

Now we can compute the final result by multiplying the expression inside the brackets by the constant outside: $$\frac{1}{2} (1) = \frac{1}{2}$$ So, the definite integral evaluates to: $$\frac{1}{2} \int_{0}^{\ln 2} e^{x} dx = \frac{1}{2}$$

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