[tex]\int_0^1 \frac{e^x}{e^x-2} d x[/tex]
Answer (2)
To evaluate the integral \int_0^1 \frac{e^x}{e^x - 2} dx, we use substitution with u = e^x - 2, changing the limits of integration accordingly. The evaluation leads to \ln(e - 2) as the final result. Thus, the answer is \ln(e - 2).
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Use substitution u = e x − 2 , so d u = e x d x .
Change the limits of integration: from x = 0 to u = − 1 , and from x = 1 to u = e − 2 .
Evaluate the integral: ∫ − 1 e − 2 u 1 d u = ln ∣ u ∣ − 1 e − 2 = ln ∣ e − 2∣ − ln ∣ − 1∣ = ln ( e − 2 ) .
The final answer is ln ( e − 2 ) .
Explanation
Problem Setup We are asked to evaluate the definite integral
∫ 0 1 e x − 2 e x d x
To do this, we will use the substitution method.
Substitution Let u = e x − 2 . Then, the derivative of u with respect to x is d u = e x d x . This substitution simplifies the integral.
Changing Limits of Integration Next, we need to change the limits of integration. When x = 0 , we have u = e 0 − 2 = 1 − 2 = − 1 . When x = 1 , we have u = e 1 − 2 = e − 2 .
Rewriting the Integral Now we can rewrite the integral in terms of u :
∫ − 1 e − 2 u 1 d u
Evaluating the Integral The integral of u 1 with respect to u is ln ∣ u ∣ . So, we have:
∫ u 1 d u = ln ∣ u ∣
Evaluating the Definite Integral Now we evaluate the definite integral:
ln ∣ u ∣ − 1 e − 2 = ln ∣ e − 2∣ − ln ∣ − 1∣ = ln ∣ e − 2∣ − ln ( 1 ) = ln ∣ e − 2∣ − 0 = ln ∣ e − 2∣
Since 2"> e ≈ 2.718 > 2 , we have 0"> e − 2 > 0 , so ∣ e − 2∣ = e − 2 . Thus, the result is ln ( e − 2 ) .
Final Answer Therefore, the final answer is ln ( e − 2 ) . We can approximate this value using a calculator: ln ( e − 2 ) ≈ − 0.33089 .
Examples
The definite integral of e x − 2 e x from 0 to 1 can be used in various applications, such as calculating the average value of a function or determining the area under a curve. For example, in physics, if e x represents the rate of a chemical reaction and e x − 2 represents a damping factor, the integral can help determine the total change in the reaction over a specific time interval. Understanding how to solve such integrals is crucial for modeling and analyzing dynamic systems in science and engineering.
