Problem 25 - College Algebra and Calculus: An Applied Approach

Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using the integration capabilities of a graphing utility. $$ \int_{0}^{9} \frac{1}{\sqrt{9-x}} d x $$

Step-by-Step Solution

Answer

The improper integral converges and its value is 6.

Step 1: Write the integral as a limit

Treat the improper integral as a limit problem. Since the function isn't defined at x=9, we replace 9 with a variable b which we will later let approach 9. This gives us: \[ \lim_{b \to 9^-}\int_{0}^{b}{\frac{1}{\sqrt{9-x}} dx}\]

Step 2: Compute the integral

Next, compute the integral by using the power rule for integration. We have:\[ \lim_{b \to 9^-} \left[-2\sqrt{9-x}\right]_0^b = \lim_{b \to 9^-} \left[-2\sqrt{9-b} - (-2\sqrt{9-0})\right]\]

Step 3: Evaluate the limit

Now, evaluate the limit: \[ \lim_{b \to 9^-} \left[-2\sqrt{9-b} +2\sqrt{9}\right] = -2\sqrt{9-9} +2\sqrt{9} = 2\sqrt{9}=6\]