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Calculus AB : Complete the Square with Integration

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Example Questions

Example Question #1 : Complete The Square With Integration

What is complete the square?

Possible Answers:

We draw the remaining sides of a square

Method for manipulating a quadratic equation for the quadratic equation

Square the function

Method for manipulating a quadratic equation into something able to be factored

Correct answer:

Method for manipulating a quadratic equation into something able to be factored

Explanation:

Say we have the equation x^2 + bx = c . It is pretty hard to solve for in this way. Instead we can use a method called complete the square. Here we will add $\frac{b}{2}^2$ to both sides which will make it factorable.

x^2 + bx = c

$x^2 +bx +\frac{b}{2}^2 = c + \frac{b}{2}^2$

$(x+\frac{b}{2})^2 = c + \frac{b}{2}^2$

$x + \frac{b}{2} = \pm \sqrt{c+\frac{b}{2}^2}$

$x = -\frac{b}{2} \pm \sqrt{c+\frac{b}{2}^2}$

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Example Question #541 : Calculus Ab

Why would one need to use complete the square when integrating?

Possible Answers:

To find the roots of a function

When we need to manipulate an equation into a recognizable identity often for a difference of squares.

Only when we are asked to in a problem

To use the quadratic formula

Correct answer:

When we need to manipulate an equation into a recognizable identity often for a difference of squares.

Explanation:

While some of these answers can be obtained by using complete the square, they are not the goal of using complete the square to integrate. When we integrate, it is often hard to see when we may need to use a trigonometric identity or a log identity. Using complete the square, we are able to manipulate our integrals in order to see these identities.

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Example Question #1 : Complete The Square With Integration

Use complete the square and show what the following equation would look like after it is applied x^2 -2x +5 = 0 .

Possible Answers:

(x-2)^2 +4 = 0

(x-1)^2 +5 = 0

x + 5 = 0

(x-1)^2 +4 = 0

Correct answer:

(x-1)^2 +4 = 0

Explanation:

To complete the square we can use one of two methods. We could use the formula x^2 + bx = c and add $\frac{b}{2}^2$ . Or if we recognize a factor that could work we can follow that route as well. So I am going to demonstrate the latter since the former is simply plugging into a formula. We will consider x^2 -2x + 5 = 0 . Now to factor our left hand side using complete the square we only consider x^2 - 2x .

x^2 - 2x looks similar to a common factorable equation x^2 - 2x + 1 . We can manipulate the equation by adding and subtracting in the same step.

x^2 - 2x + 1 - 1 = (x-1)^2 -1

We plug this back into our original equation:

x^2 - 2x +5 =0

(x-1)^2 - 1 +5 = 0

(x-1)^2 +4 = 0

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Example Question #2 : Complete The Square With Integration

Show what the following equation will look like after completing the square. 0 = 1 - x^2 - 6x

Possible Answers:

0 = 1 - (x+3)^2

0 = 10 - (x+9)^2

0 = 10 - (x-3)^2

0 = 10 - (x+3)^2

Correct answer:

0 = 10 - (x+3)^2

Explanation:

In order to complete the square recognize that x^2 + 6x + 9 = (x + 3)^2 . We proceed as follows:

0 = 1 - x^2 - 6x

0 = 1 -(x^2 + 6x)

0 = 1 - (x^2 + 6x + 9 - 9)

0 = 10 - (x+3)^2

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Example Question #3 : Complete The Square With Integration

Evaluate the following integral using complete the square.

$\int \frac{1}{x^2 + 10x +26}dx$

Possible Answers:

arccos(x+5) + C

arcsin(x+5) + C

arctan(x+5) + C

arcsec(x+5) + C

Correct answer:

arctan(x+5) + C

Explanation:

In order to complete the square for the quadratic we first consider the denominator of the integral function.

x^2 +10x + 26

x^2 + 10x + 25 - 25 + 26

(x + 5)^2 +1

Now our integral looks like $\int \frac{1}{(x+5)^2 +1} dx$ . We will use a u substitutions where u = x+5 . Now we have the integral:

$\int\frac{1}{u^2+1}dx$ .

We know that the integral $\int\frac{dx}{x^2 + 1} = arctan(x) + C$ . So our integral proceeds as follows:

$\int\frac{1}{u^2 + 1}dx = arctan(u) + C = arctan(x+5) + C$ .

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Example Question #1 : Complete The Square With Integration

What would be the proper difference of squares to use to integrate the following to be equal to $\int\frac{dx}{1 - x^2 - 4x} = arcsin(\frac{x+2}{\sqrt{5}}) + C$ ?

Possible Answers:

$\sqrt(5)^2 - (x+2)^2$

1 - (x+2)^2

5^2 - (x+2)^2

4 - (x+2)^2

Correct answer:

$\sqrt(5)^2 - (x+2)^2$

Explanation:

If we want to integrate the following to equal $arcsin(\frac{x+2}{\sqrt{5}}) +C$ we first must recall that $\int\frac{dx}{a^2 - x^2} = arcsin(\frac{x}{a}) + c$ . We must consider the denominator the integral:

1 - x^2 - 4x

1 - (x^2 +4x)

1 -(x^2 + 4x +4 - 4)

5 -(x^2 + 4x + 4)

5 -(x + 2)^2

Now we use a u substitution where u = x+2

5 - u^2

We also know that $\sqrt(5)^2 = 5$ so we can substitute that in as well

$\sqrt(5)^2 - u^2$

If we plug this back into our integral we can now solve this integral easily.

$\int\frac{dx}{\sqrt(5)^2 - u^2} = arcsin(u) + C = arcsin(\frac{x+2}{\sqrt{5}}) + C$ . The difference of two squares that we needed was $\sqrt(5) - (x+2)^2$

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Example Question #5 : Complete The Square With Integration

Use complete the square to solve the following integral

$\int\frac{dx}{x^2+6x+7}$

Possible Answers:

$ln|(x+3) + \sqrt{(x+3)^2 - \sqrt{2}} + C$

$arctan(\frac{(x+3)^2 }{\sqrt{2}}) + C$

$arcsin(\frac{(x+3)^2 }{\sqrt{2}}) + C$

$ln|(x+3) + \sqrt{(x+3)^2 - 2} + C$

Correct answer:

$ln|(x+3) + \sqrt{(x+3)^2 - 2} + C$

Explanation:

To solve this integral, we must first complete the square of the denominator.

x^2 + 6x + 7

x^2 + 6x + 9 - 9 +7

(x+3)^2 - 2

Plugging this back into our integral we have $\int\frac{dx}{(x+3)^2 - 2}$ . Recall that $\int\frac{dx}{x^2 \pm a^2} = ln|x + \sqrt{x^2 \pm a^2}|$

$\int\frac{dx}{(x+3)^2 - 2}$

$\int\frac{dx}{(x+3)^2 - \sqrt{2}^2}$

Using a u substitution we will let u = x+2

$\int\frac{dx}{u^2 - \sqrt{2}^2} = ln|u + \sqrt{u^2 - \sqrt{2}^2}| + C = ln|(x+3) + \sqrt{(x+3)^2 - 2}| + C$

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Example Question #6 : Complete The Square With Integration

Use complete the square to solve the following integral

$\int\frac{dx}{-3 -x^2 -4x}$

Possible Answers:

$\frac{1}{2}ln|\frac{3 + x}{x-1}| + C$

$\frac{1}{2}ln|\frac{x-1}{3+x}| + C$

$ln|\frac{3 + x}{x-1}| + C$

$\frac{1}{2}ln|\frac{3 - x}{x+1}| + C$

Correct answer:

$\frac{1}{2}ln|\frac{3 + x}{x-1}| + C$

Explanation:

We start by considering the denominator.

-3 - x^2-4x

-3 - (x^2 +4x)

-3 - (x^2 +4x +4 -4)

1 - (x^2 +4x +4)

1 - (x+2)^2

Plugging this back into our integral we get $\int\frac{dx}{1-(x+2)^2$ . Using a u substitution, we will let u = x+2 .

$\int\frac{dx}{1 - u^2}$

Recall that $\int\frac{dx}{a^2 - x^2} = \frac{1}{2a}ln|\frac{a+x}{a-x}|+C$

$\int\frac{dx}{1 - u^2}$ $= \frac{1}{2}ln|\frac{1+u}{1-u}| +C = \frac{1}{2}ln|\frac{3 + x}{x-1} + C$

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Example Question #1 : Complete The Square With Integration

True or False: We can only use complete the square if it gives us an exact formula for a solution to the integral.

Possible Answers:

False

True

Correct answer:

False

Explanation:

This is not true. We use complete the square in order to get the integral into a familiar form but often times we need to use a u substitution in order to be able to complete the problem.

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Example Question #2 : Complete The Square With Integration

True or False: We are able to complete the square by adding and subtracting the same term to our given equation. This is basically adding zero, and therefore it does not change the value of our equation.

Possible Answers:

True

False

Correct answer:

True

Explanation:

Take x^2 -2x +5 = 0 for example. We are able to complete the square by adding and subtracting . When we do this, it does not change the value of our equation and this is because of and are identities of each other meaning that 1 + (-1) = 0 . By doing this with any number (adding and subtracting the same number) we are able to successfully complete the square without changing the value of our equation.

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Calculus AB : Complete the Square with Integration

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #1 : Complete The Square With Integration

Example Question #541 : Calculus Ab

Example Question #1 : Complete The Square With Integration

Example Question #2 : Complete The Square With Integration

Example Question #3 : Complete The Square With Integration

Example Question #1 : Complete The Square With Integration

Example Question #5 : Complete The Square With Integration

Example Question #6 : Complete The Square With Integration

Example Question #1 : Complete The Square With Integration

Example Question #2 : Complete The Square With Integration

All Calculus AB Resources

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