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Chapter 17: Problem 2

Find \(\int_{0}^{\infty} \mathrm{e}^{-2 x} \mathrm{~d} x\)

Short Answer

Expert verified
Question: Evaluate the improper integral \(\int_{0}^{\infty} \mathrm{e}^{-2 x} \mathrm{d} x\). Answer: \(\frac{1}{2}\)

Step by step solution

01

Identify the function to be integrated

The function to be integrated in this problem is e^{-2x}.
02

Compute the integral

Start by finding the antiderivative of this function. The antiderivative of e^{-2x} can be computed using standard rules from calculus, in this case, the power rule of integration. The power rule of integration states that the integral of e^{ax} with respect to x is \frac{1}{a}e^{ax}. Applying this rule with a = -2, we get the antiderivative as -\frac{1}{2}e^{-2x}.
03

Evaluate the integral from 0 to infinity

Now, this antiderivative function has to be evaluated from 0 to infinity. Technically, this involves taking the limit as x approaches infinity. So, the integral from 0 to infinity of e^{-2x} dx is: \[ - \frac{1}{2} \left[\lim_{x \to \infty}e^{-2x} - e^{0} \right] \]
04

Substitute the limits

Substitute the limits into this equation. As the first term is a limit as x approaches infinity for the function e^{-2x}, it will reach zero (since exponential of negative large number will become closer to zero). The second term becomes 1 when the zero is substituted for x (since any number to the power of 0 is 1). So, \[ - \frac{1}{2} \left[\lim_{x \to \infty}e^{-2x} - e^{0} \right] = - \frac{1}{2} \left[0 - 1 \right] \]
05

Finalize the solution

Now solving the above equation, we get result as \frac{1}{2}. That is, \(\int_{0}^{\infty} \mathrm{e}^{-2 x} \mathrm{d} x = \frac{1}{2}\).

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