Chapter 17: Problem 17

Find $\int_{0}^{\infty} \mathrm{e}^{-2 t} \mathrm{~d} t$.

Short Answer

Expert verified

Question: Evaluate the improper integral: $\int_{0}^{\infty} e^{-2t} dt$ Answer: $\frac{1}{2}$

Step by step solution

Find the indefinite integral of $e^{-2t}$

To find the indefinite integral of the given function, we need to find a function that has $e^{-2t}$ as its derivative. Recall that the derivative of $e^{ax}$ (where $a$ is a constant) is $ae^{ax}$. Therefore, the integral of $e^{-2t}$ with respect to $t$ is: $$-\frac{1}{2}e^{-2t} + C$$ where $C$ is the constant of integration.

Find the definite integral using limits

Now, we need to find the definite integral of $e^{-2t}$ from $0$ to $\infty$. Since we're dealing with an improper integral, we'll use limits to evaluate it: $$\int_{0}^{\infty} e^{-2t} dt = \lim_{b \to \infty} \int_{0}^{b} e^{-2t} dt.$$ Now, we'll use the antiderivative we found in Step 1 to evaluate the above expression: $$\lim_{b \to \infty} \left[-\frac{1}{2}e^{-2t}\right]_0^b = \lim_{b \to \infty} \left[-\frac{1}{2}e^{-2b} - \left(-\frac{1}{2}e^0\right)\right].$$

Evaluate the limit

Finally, we'll evaluate the limit as $b$ goes to $\infty$: $$\lim_{b \to \infty} \left[-\frac{1}{2}e^{-2b} - \left(-\frac{1}{2}e^0\right)\right] = -\frac{1}{2}\lim_{b \to \infty} e^{-2b} + \frac{1}{2}.$$ Since the exponential function converges to $0$ as its argument goes to $-\infty$, we have: $$\lim_{b \to \infty} e^{-2b} = 0.$$ Therefore, the value of the definite integral is: $$-\frac{1}{2}\cdot 0 + \frac{1}{2} = \frac{1}{2}.$$

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

Short Answer

Step by step solution

Find the indefinite integral of \(e^{-2t}\)

Find the definite integral using limits

Evaluate the limit

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