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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

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Example Questions

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 #1 : 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:

False

True

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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Example Question #1 : Model And Verify Differential Equations

What is a differential equation?

Possible Answers:

A function that has both positive and negative intervals in its domain

An equation that is completely negative

An equation with a function and a least one of its derivatives

An equation in which we are taking the difference of two functions

Correct answer:

An equation with a function and a least one of its derivatives

Explanation:

Let’s say we have a function and that function plus its derivative is a solution to our function f(x) = 5x . This gives us the differential equation:

$y + \frac{dy}{dx} = 5x$

This is a differential equation because it has both the function and its derivative as part of the equation.

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Example Question #1 : Model And Verify Differential Equations

Which of the following is a differential equation?

Possible Answers:

f(x) = y

$f(x) = \frac{3}{2}x$

$f(x) = \frac{d}{dx} + 3cos(\theta)$

f(x) = y + x

Correct answer:

$f(x) = \frac{d}{dx} + 3cos(\theta)$

Explanation:

Recall that a differential equation is an equation that has both the function and at least one of its derivatives in the equation. The last answer, $f(x) = \frac{d}{dx} + 3cos(\theta)$ , is the only equation that has both the function and the derivative in the equation so this is our answer.

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Example Question #3 : Model And Verify Differential Equations

Is the equation y = 2x a solution to the differential equation $\frac{dy}{dx} = \frac{y}{2x}$ ?

Possible Answers:

Yes

There is not enough information

Correct answer:

Explanation:

First we need to take the derivative with respect to of our function .

y = 2x

$\frac{dy}{dx} = 2$

Now we can plug this back into our differential equation that we were given.

$\frac{dy}{dx} = \frac{y}{2x}$

$2 = \frac{y}{2x}$

To simplify things further, let’s also write in terms of . So our original function says y = 2x . We will plug in for all terms in the differential equation.

$2 = \frac{y}{2x}$

$2 = \frac{2x}{2x}$

$2 \neq 1$

And so y = 2x is not a solution to the differential equation $\frac{dy}{dx} = \frac{y}{2x}$ .

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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

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 #1 : Complete The Square With Integration

Example Question #1 : Model And Verify Differential Equations

Example Question #1 : Model And Verify Differential Equations

Example Question #3 : Model And Verify Differential Equations

All Calculus AB Resources

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