Problem 33
Question
Completing the Square In Exercises \(33-42,\) find or evaluate the integral by completing the square. $$ \int_{0}^{2} \frac{d x}{x^{2}-2 x+2} $$
Step-by-Step Solution
Verified Answer
\(\frac{\pi}{2}\)
1Step 1: Completing the Square
We have \(x^2 - 2x + 2\) as the denominator of our integral. We can write it as \((x - 1)^2 + 1\) using the formula \(a^2 - 2ab + b^2\) which is a perfect square, where \(a = x\) and \(b = 1\).
2Step 2: Substitution
Next, perform a substitution to get the integral into a standard form. Let \(u = x - 1\), then \(du = dx\). The integral becomes: \(\int \frac{du}{u^2 + 1}\).
3Step 3: Finding the Antiderivative
This integral now has the standard form \(\int \frac{du}{u^2 + 1}\), whose antiderivative is \(\tan^{-1}(u) + C\).
4Step 4: Substituting back and Finding the Definite Integral
First, the \(u\) should be replaced by \(x - 1\), to give us \(\tan^{-1}(x - 1) + C\). Secondly, place the original limits from \(0\) to \(2\), giving us: \[\tan^{-1}(2 - 1) - \tan^{-1}(0 - 1) = \tan^{-1}(1) - \tan^{-1}(-1)\] which simplifies to \(\frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2}\).
Key Concepts
Definite IntegralSubstitution MethodAntiderivativeTrigonometric Integration
Definite Integral
A definite integral is a key concept in calculus used to find the area under a curve over a specified range. Unlike indefinite integrals, which represent a family of functions, definite integrals yield a numerical value.
In our problem, we evaluate the integral from 0 to 2, meaning we look for the area under the curve defined by the function \ \( \frac{1}{x^2 - 2x + 2} \ \) between these two limits.
This process provides the net area, taking into account any portions of the curve that fall below the x-axis.
In our problem, we evaluate the integral from 0 to 2, meaning we look for the area under the curve defined by the function \ \( \frac{1}{x^2 - 2x + 2} \ \) between these two limits.
- The endpoints are denoted by the numbers 0 and 2 at the integral's bounds.
- You're not just finding an expression, but an exact numeric value from a specific interval.
This process provides the net area, taking into account any portions of the curve that fall below the x-axis.
Substitution Method
The substitution method is a powerful and commonly used technique to simplify complex integrals. It is essentially a change of variable that makes the integral easier to evaluate.
- Identify a part of the integrand complicating the process, often something squared or under a square root.
- Replace it with a simpler variable, say \(u\). For our integral, \(x - 1\) was replaced with \(u\).
- Determine the corresponding derivative. Here, \(du = dx\).
Antiderivative
Finding the antiderivative is like working backwards from a derivative. It's the process of calculating the function whose derivative is the function you're integrating.
For the integral \ \( \int \frac{du}{u^2 + 1} \ \), we know that its antiderivative is \( \tan^{-1}(u) + C \), where \(C\) is the constant of integration. This step is crucial because:
For the integral \ \( \int \frac{du}{u^2 + 1} \ \), we know that its antiderivative is \( \tan^{-1}(u) + C \), where \(C\) is the constant of integration. This step is crucial because:
- It converts our simplified integrand back into a function.
- If it's a definite integral like in our problem, you evaluate this antiderivative within the specified limits (between 0 and 2 in this case).
Trigonometric Integration
Trigonometric integration typically involves integrals where the integrand is a trigonometric function. In our case, the problem transitions into trigonometric integration once the substitution leads to the integral \ \( \int \frac{du}{u^2 + 1} \ \).
- The function \( \frac{1}{u^2 + 1} \) is commonly encountered and directly related to the arctangent function, \( \tan^{-1}(u) \).
- This highlights the importance of recognizing trigonometric identities and their derivatives which assist in solving these integrals.
Other exercises in this chapter
Problem 32
Show that \(f\) is strictly monotonic on the given interval and therefore has an inverse function on that interval. \(f(x)=\cot x, \quad(0, \pi)\)
View solution Problem 33
In Exercises 33–36, find an equation of the tangent line to the graph of the function at the given point. $$ y=\sinh \left(1-x^{2}\right), \quad(1,0) $$
View solution Problem 33
Solving an Equation In Exercises \(25-34,\) solve the equation accurate to three decimal places. $$ \log _{3} x^{2}=4.5 $$
View solution Problem 33
Solve the equation for \(x\). \(\arcsin (3 x-\pi)=\frac{1}{2}\)
View solution