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Chapter 5: Problem 33

Completing the Square In Exercises \(33-42,\) find or evaluate the integral by completing the square. $$ \int_{0}^{2} \frac{d x}{x^{2}-2 x+2} $$

Short Answer

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

Step by step solution

01

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 \(b = 1\).
02

Substitution

Next, perform a substitution to get the integral into a standard form. Let \(u = x - 1\), then \(du = dx\). The integral becomes: \(\int \frac{du}{u^2 + 1}\).
03

Finding the Antiderivative

This integral now has the standard form \(\int \frac{du}{u^2 + 1}\), whose antiderivative is \(\tan^{-1}(u) + C\).
04

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 \(\frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.
  • 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.
Once you have performed all the necessary steps of solving the integrand and finding an antiderivative, you apply the limits of integration.

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, \(du = dx\).
The substitution redefines the integral in terms of \(u\), turning the integral into \ \( \int \frac{du}{u^2 + 1} \ \). As a result, it matches recognizable standard forms, paving the way for easier integration.
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:
  • 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).
Once known, we substitute back the original variable to express our solution in the initial context of the problem.
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.
After performing the integration, you take advantage of your known trigonometric functions to make sense within the problem's context. This results in the calculated value required by the definite integral. Here, it simplifies down to an expression involving \( \tan^{-1} \), resulting eventually in values like \(\frac{\pi}{4} - (-\frac{\pi}{4})\), producing a final result of \(\frac{\pi}{2}\).

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Most popular questions from this chapter

Use a computer algebra system to find the linear approximation \(P_{1}(x)=f(a)+f^{\prime}(a)(x-a)\) and the quadratic approximation \(P_{2}(x)=f(a)+f^{\prime}(a)(x-a)+\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}\) of the function \(f\) at \(x=a\) . Sketch the graph of the function and its linear and quadratic approximations. \(f(x)=\arctan x, \quad a=1\)

Find the derivative of the function. \(y=\frac{1}{2}\left[x \sqrt{4-x^{2}}+4 \arcsin \left(\frac{x}{2}\right)\right]\)

Find any relative extrema of the function. \(f(x)=\operatorname{arcsec} x-x\)

Area In Exercises \(125-128\) , find the area of the region bounded by the graphs of the equations. Use a graphing utility to graph the region and verify your result. $$ y=e^{-2 x}+2, y=0, x=0, x=2 $$

In Exercises 87鈥�90, solve the differential equation. $$ \frac{d y}{d x}=\frac{x^{3}-21 x}{5+4 x-x^{2}} $$
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