Problem 24 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Either evaluate the given improper integral or show that it diverges. $$ \int_{0}^{+\infty} 3 e^{-5 x} d x $$

Step-by-Step Solution

Answer

Since \[ \lim_{b \rightarrow \infty} -\frac{3}{5} e^{-5b} + \frac{3}{5} = \frac{3}{5} \], The improper integral converges to \frac{3}{5}.

Step 1 - Identify the Improper Integral

An improper integral is given by \ \ \ \ \[ \int_{0}^{+\infty} 3 e^{-5 x} d x \] \ The integral is improper because it involves an infinite upper limit.

Step 2 - Substitute the Limit

To evaluate the improper integral, substitute the infinite limit with a variable and consider the limit as this variable approaches infinity. Let \[\b \rightarrow \infty\] then evaluate \ \ \ \ \[ \int_{0}^{b} 3 e^{-5 x} d x \] \ and take the limit as \[ \b \rightarrow \infty \].

Step 3 - Evaluate the Definite Integral

Integrate the function inside the integral: \ \ \ \ \[ \int 3 e^{-5 x} d x = -\frac{3}{5} e^{-5 x} + C \] \ Evaluate this antiderivative from 0 to \[ \b \] : \ \ \ \[ \left[ -\frac{3}{5} e^{-5 x} \right]_{0}^{b} = -\frac{3}{5} e^{-5 \b} + \frac{3}{5} \]

Step 4 - Apply the Limit

Take the limit of the result as \[ b \rightarrow \infty \]: \ \ \ \[ \lim_{b \rightarrow \infty} \left( -\frac{3}{5} e^{-5 b} + \frac{3}{5} \right) \] when \[ b \rightarrow \infty \] , \[ e^{-5 b} \rightarrow 0 \] . So: \ \[ \lim_{b \rightarrow \infty} \left( -\frac{3}{5} e^{-5 b} + \frac{3}{5} \right) = \frac{3}{5} \]

Key Concepts

Definite Integral

A definite integral calculates the area under a curve within a specific interval, which is defined by the limits of integration. In our original problem, we dealt with an integral from 0 to \(\b\). Here, \(\b\) is a placeholder that we later let approach infinity to match the upper limit of the improper integral. The formula for a definite integral is \(\bigg[ F(x) \bigg]_a^b = F(b) - F(a)\). This means you find the antiderivative of the function you are integrating and then subtract the antiderivative evaluated at the lower limit from the antiderivative evaluated at the upper limit.

Limit at Infinity

Limits at infinity help us understand the behavior of functions as the input grows very large. When evaluating our integral, we dealt with \(\b \rightarrow \)∞. This means we substitute \(\b\) with a very large number and see what happens to the expression. For the exponential function \( e^{-5 b}\), we observe that it approaches 0 as \( b \rightarrow \)∞ because the exponent is increasingly negative. The key property here is that \( e^{-k x}\) approaches 0 as \( x \rightarrow \)∞ for any positive constant \( k\).

Exponential Function

The exponential function is crucial in our integral. Specifically, we handle \( 3 e^{-5 x} \). Exponential functions have the form \( e^{kx} \) where \( k \) determines the growth or decay rate. In our problem, since \( k = -5 \), the function \( 3 e^{-5 x} \) decays quickly as \( x \) increases. Because the function decreases rapidly, the area under the curve from 0 to infinity (our integral) is finite. This is why we were able to find a finite result, \(\frac{3}{5}\), when solving the improper integral.