Problem 35 Complete the square and give a s... [FREE SOLUTION]

91影视

Calculus

Deborah Hughes-Hallett, Andrew M. Gleason, William G. McCallum

$Math Studyset 91影视 Explanations$ Math

7 Edition

Chapter 7: Problem 35

Complete the square and give a substitution (not necessarily trigonometric) which could be used to compute the integrals.$$\int \frac{z-1}{\sqrt{2 z-z^{2}}} d z$$.

Short Answer

Expert verified

Use substitution $u = z - 1$ to evaluate $\int \frac{u}{\sqrt{1-u^2}} \, du$.

Step by step solution

Identify the quadratic expression under the square root

The expression under the square root in the integrand is $2z - z^2$. We need to rewrite this quadratic expression in a form that will allow us to complete the square.

Rearrange the quadratic expression

Write the quadratic $2z - z^2$ in the standard form $-z^2 + 2z$. This helps us to see the terms better for completing the square.

Factor out the negative

Factor $-1$ from the quadratic terms: $-z^2 + 2z = -(z^2 - 2z)$. Now, we can focus on completing the square inside the parentheses.

Complete the square

To complete the square, take half of the coefficient of $z$ from $z^2 - 2z$, which is $-2$, giving $-1$, and then square it, which results in $1$. Thus, $z^2 - 2z = (z-1)^2 - 1$. Now, substitute back: $-((z-1)^2 - 1) = -(z-1)^2 + 1$.

Re-expression of the original integral

Substitute back this completed square into the original integral: $\int \frac{z-1}{\sqrt{-(z-1)^2 + 1}} \, dz$. This represents $\int \frac{z-1}{\sqrt{1-(z-1)^2}} \, dz$.

Determine substitution for the integral

Recognizing the form $\sqrt{1-(z-1)^2}$ suggests a trigonometric substitution, but to avoid it as per the instruction, use $u = z - 1$. This simplifies the integral to $\int \frac{u}{\sqrt{1-u^2}} \, du$ by the substitution $z = u+1$, $dz = du$.

Simplifying the integral

Now the integral becomes $\int \frac{u}{\sqrt{1-u^2}} \, du$, which can be evaluated with a known antiderivative or further substitution if necessary.

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Key Concepts

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

Quadratic Expressions

Quadratic expressions are algebraic expressions that involve terms having a degree of two. In simpler terms, a quadratic expression takes the form $ax^2 + bx + c$, where $a$, $b$, and $c$ are constants, and $a eq 0$. These expressions can often be found in various mathematical scenarios, including integrals.

The quadratic expression we encountered in the exercise is $2z - z^2$.
Our task was to rewrite it into a form using completing the square technique, helping us simplify the integration process.

By completing the square, we can transform the quadratic expression to a perfect square form, making it easier to integrate or solve within different contexts such as finding roots or solving inequalities. In this context, rewriting the expression $2z - z^2$ as $-(z-1)^2 + 1$ allows us to reformulate and solve the integral more effectively.

Integral Substitution

Integral substitution is a critical technique used in calculus to simplify and evaluate integrals. It involves substituting part of the integral with a new variable, thereby changing the integral into a simpler form. This technique is particularly useful when you notice specific patterns within the integrand.

For the given integral $\int \frac{z-1}{\sqrt{2z-z^{2}}} \, dz$, we employed substitution after completing the square.
We used the substitution $u = z - 1$ to rewrite and simplify the integral into $\int \frac{u}{\sqrt{1-u^2}} \, du$.

This adjusted form is easier to handle and can be evaluated using known antiderivatives or additional techniques. Such substitutions are powerful with expressions that resemble standard calculus identities, helping in unraveling complex integrals seamlessly.

Integration Techniques

Integration techniques are methods that help us solve integrals that are not straightforward. One powerful technique involves transformation or re-expression of integrals into forms that are simpler to work with.

In our exercise, we saw how completing the square allowed for a crucial transformation.
After substitution, the integral $\int \frac{u}{\sqrt{1-u^2}} \, du$ became more manageable.

The process may also involve recognizing known patterns such as trigonometric integrals. For instance, an integral of the form $\int \frac{u}{\sqrt{1-u^2}} \, du$ is aligned with standard integrals involving arcsine or arccosine functions. By leveraging these techniques, we can effectively address integrals that initially appear complex, streamlining the problem-solving process in calculus.

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