Problem 23 - Essential Calculus

Question

Find or evaluate the integral. (Complete the square, if necessary.) $$ \int_{0}^{2} \frac{d x}{x^{2}-2 x+2} $$

Step-by-Step Solution

Answer

\(\frac{\pi}{2}\)

Step 1: Rewrite the function

Rewrite the integral in a simpler form by completing the square in the denominator:\[ \int_{0}^{2} \frac{d x}{x^{2}-2 x+2} = \int_{0}^{2} \frac{dx}{(x-1)^2+1}\]

Step 2: Substitute variables

By using the substitution \(u = x - 1\), find the new limits of integration \(u_{0} = -1\) and \(u_{2} = 1\), and express the integral as follows:\[ \int_{-1}^{1} \frac{du}{u^2+1}\]

Step 3: Evaluate integral

Now evaluate this simpler integral which is a standard form that is memorized / known to be \( \arctan(u) \):\[ \= \left.\arctan(u)\right|_{-1}^{1} = \arctan(1) - \arctan(-1)\]

Step 4: Evaluating Limits

Finally, evaluate the definite integral by substituting the new limits of \(u\) into the antiderivative: \[= \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}\]