Problem 8 - Calculus Early Transcendentals

Question

Evaluate the following integrals or state that they diverge. $$\int_{1}^{\infty} 2^{-x} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the improper integral, or state whether it diverges: $$\int_{1}^{\infty} 2^{-x} d x$$ Answer: The integral converges, and its value is: $$\int_{1}^{\infty} 2^{-x} d x = \frac{1}{2\ln{2}}$$

Step 1: Transform the integral into a limit

Since the integral has an infinite limit, it's considered as an improper integral. We need to transform it into a limit in order to evaluate it. We have: $$\int_{1}^{\infty} 2^{-x} d x = \lim_{a \to \infty} \int_{1}^{a} 2^{-x} d x$$

Step 2: Find the antiderivative of the function

Next, we need to find the antiderivative of the function 2^{-x}. To do this, we first need to rewrite the function as an exponential function with base e: $$2^{-x} = e^{-x \cdot \ln{2}}$$ Now, we find the antiderivative: $$\int e^{-x \cdot \ln{2}} d x = -\frac{1}{\ln{2}} \int e^{-x \cdot \ln{2}} (\ln{2}) d x$$ Applying the substitution rule with \(u = -x \cdot \ln{2}\) and a change of variables, we get: $$-\frac{1}{\ln{2}} \int e^{u} d\left(\frac{u}{-\ln{2}}\right) = -\frac{1}{\ln{2}} e^{u} = -\frac{1}{\ln{2}} e^{-x \cdot \ln{2}}$$ Hence, the antiderivative of the function 2^{-x} is: $$F(x) = -\frac{1}{\ln{2}} e^{-x \cdot \ln{2}} + C$$

Step 3: Evaluate the limit

Now, we will use the antiderivative to evaluate the limit at the bounds of the integral: $$\lim_{a \to \infty} \int_{1}^{a} 2^{-x} d x = \lim_{a \to \infty} \left[ -\frac{1}{\ln{2}} e^{-x \cdot \ln{2}}\right]_1^a$$ Now we plug in the values of the limits and evaluate the limit as \(a\) goes to infinity: $$\lim_{a \to \infty} \left[-\frac{1}{\ln{2}} e^{-a \cdot \ln{2}} - \left(-\frac{1}{\ln{2}} e^{-1 \cdot \ln{2}}\right)\right]$$ $$= \lim_{a \to \infty} \left[-\frac{1}{\ln{2}} e^{-a \cdot \ln{2}} + \frac{1}{\ln{2}} e^{-\ln{2}}\right]$$ As \(a\) goes to infinity, the exponential term \(e^{-a \cdot \ln{2}}\) goes to 0 because it has a negative exponent. So, the expression becomes: $$-\frac{1}{\ln{2}}( 0) + \frac{1}{\ln{2}}( \frac{1}{2}) = \frac{1}{2\ln{2}}$$ Hence, the integral converges, and its value is: $$\int_{1}^{\infty} 2^{-x} d x = \frac{1}{2\ln{2}}$$