Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{0}^{\infty} \frac{5}{e^{2 x}} d x $$

Short answer

The improper integral \[\int_{0}^{\infty} \frac{5}{e^{2 x}} d x\] converges and its value is \(\frac{5}{2}\).

Step-by-step solution

Rewrite the Integral

To simplify the integral, it is rewritten as: \[\int_{0}^{\infty} \frac{5}{e^{2 x}} d x = 5\int_{0}^{\infty}e^{-2x} dx\]

Define the Improper Integral

The integral from 0 to infinity is a type of improper integral. It's defined as a limit: \[5\int_{0}^{\infty}e^{-2x} dx = 5\lim_{b \to \infty} \int_{0}^{b} e^{-2x} dx\]

Evaluate the Integral

Evaluate the integral using the antiderivative of the function \(e^{-2x}\), which is \(-\frac{1}{2}e^{-2x}\):\[5\lim_{b \to \infty}[ -\frac{1}{2}e^{-2x}\Big|_0^b]\]\Simplify to get: \[5\lim_{b \to \infty}[ -\frac{1}{2}e^{-2b}+\frac{1}{2}]\]

Evaluate the Limit

Now, calculate the limit as b approaches infinity: \[5*(-\frac{1}{2}*0+\frac{1}{2}) = 5*\frac{1}{2} = \frac{5}{2}\] As the limit is finite, the integral converges and its value is \(\frac{5}{2}\).