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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

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45 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Use Substitution And Integration

Solve the following integral using substitution:

$\int{x^3(2+x^4)^5}dx$

Possible Answers:

$\frac{1}{24}(2+x^4)^6+c$

$2x+x^{20}+c$

$2x^3+x^{23}+c$

$\frac{1}{6}(2+x^4)^6+c$

$\frac{1}{2}x^4+\frac{1}{24}x^{24}+c$

Correct answer:

$\frac{1}{24}(2+x^4)^6+c$

Explanation:

To solve the integral, we have to simplify it by using a variable u to substitute for a variable of x.

For this problem, we will let u replace the expression 2+x^4 .

Next, we must take the derivative of u. Its derivative is du=4x^3dx .

Next, solve this equation for dx so that we may replace it in the integral.

Plug in place of (2+x^4) and $\frac{du}{4x^3}$ in place of into the original integral and simplify.

The x^3 in the denominator cancels out the remaining x^3 in the integral, leaving behind a $\frac{1}{4}$ . We can pull the $\frac{1}{4}$ out front of the integral. Next, take the anti-derivative of the integrand and replace u with the original expression, adding the constant to the answer.

The specific steps are as follows:

1. $\int{x^3(2+x^4)^5dx}$

2. u=2+x^4

3. du=4x^3dx

4. $dx=\frac{du}{4x^3}$

5. $\int{x^3u^5(\frac{du}{4x^3})}$

6. $\frac{1}{4}\int{u^5}du$

7. $\frac{1}{4}(\frac{1}{6}u^6)$

8. $\frac{1}{4}(\frac{1}{6}(2+x^4)^6$

9. $\frac{1}{24}(2+x^4)^6+c$

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Example Question #2 : Use Substitution And Integration

Solve the following integral using substitution:

$\int{x^2\sqrt{x^3+1}}dx$

Possible Answers:

$\frac{2}{3}(x^3+1)^9+c$

$\frac{1}{3}(x^3+1)^{\frac{5}{2}}+c$

$x^{\frac{7}{2}}+x^2+c$

$\sqrt[3]{(x^3+1)}(\frac{1}{x^2})+c$

$\frac{2}{9}(x^3+1)^{\frac{3}{2}}+c$

Correct answer:

$\frac{2}{9}(x^3+1)^{\frac{3}{2}}+c$

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression x^3+1 . Next, we must take the derivative of u. Its derivative is 3x^2dx . Next, solve this equation for so that we may replace it in the integral. Plug in place of x^3+1 and $\frac{du}{3x^2}$ in place of into the original integral and simplify. The x^2 in the denominator cancels out the remaining x^2 in the integral, leaving behind a $\frac{1}{3}$ . We can pull the $\frac{1}{3}$ out front of the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{x^2\sqrt{x^3+1}}dx$

2. = x^3+1

3. du=3x^2dx

4. $\int{x^2\sqrt{u}\frac{du}{3x^2}}$

5. $\frac{1}{3}\int{\sqrt{u}du}$

6. $\frac{1}{3}(\frac{2}{3}u^{\frac{3}{2}})+c$

7. $\frac{2}{9}u^{\frac{3}{2}}+c$

8. $\frac{2}{9}(x^3+1)^{\frac{3}{2}}+c$

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Example Question #3 : Use Substitution And Integration

Solve the following integral using substitution:

$\int{cos^3xsinxdx}$

Possible Answers:

$\frac{-1}{3}(cosx)^3+c$

$\frac{1}{3}(cosx)^3+c$

cotxcos^2x+c

$\frac{-cos^4x}{4}+c$

$\frac{sin^4x}{4}+c$

Correct answer:

$\frac{-cos^4x}{4}+c$

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression cosx . Next, we must take the derivative of . Its derivative is -sinxdx . Next, solve this equation for so that we may replace it in the integral. Plug in place of cosx and $\frac{du}{-sinx}$ in place of into the original integral and simplify. The sinx in the denominator cancels out the remaining sinx in the integral, leaving behind a . Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{cos^3xsinxdx}$

2. u=cosx

3. du=-sinxdx

4. $\int{u^3sinx(\frac{du}{-sinx})}$

5. $\int-u^3du$

6. $\frac{-1}{4}u^4+c$

7. $\frac{-cos^4x}{4}+c$

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Example Question #4 : Use Substitution And Integration

Solve the following integral using substitution:

$\int{xsin(x^2)dx}$

Possible Answers:

$\frac{1}{2}cos(x^2)+c$

2cos(x^2)+c

$\frac{-1}{2}cos(x^2)+c$

$\frac{-1}{2}sin(x^2)+c$

$\frac{1}{2}sin(x^2)+c$

Correct answer:

$\frac{-1}{2}cos(x^2)+c$

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression x^2 . Next, we must take the derivative of . Its derivative is . Next, solve this equation for so that we may replace it in the integral. Plug in place of x^2 and $\frac{du}{2x}$ in place of into the original integral and simplify. The in the denominator cancels out the remaining in the integral, leaving behind a $\frac{1}{2}$ . We can pull the $\frac{1}{2}$ out front of the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{xsin(x^2)dx}$

2. u=x^2

3. du=2xdx

4. $\int{x(sinu)\frac{du}{2x}}$

5. $\frac{1}{2}\int{sinudu}$

6. $\frac{-1}{2}cosu+c$

7. $\frac{-1}{2}cos(x^2)+c$

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Example Question #5 : Use Substitution And Integration

Solve the following integral using substitution:

$\int{x^2e^{x^3}dx}$

Possible Answers:

$\frac{3}{2}e^{x^3}+c$

$\frac{1}{3}e^{x^3}+c$

$2xe^{x^3}+c$

$\frac{1}{2}e^{x2}+c$

$\frac{3}{2}e^{x^2}+c$

Correct answer:

$\frac{1}{3}e^{x^3}+c$

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression x^3 . Next, we must take the derivative of . Its derivative is 3x^2 . Next, solve this equation for so that we may replace it in the integral. Plug in place of x^3 and $\frac{du}{3x^2}$ in place of into the original integral and simplify. The x^2 in the denominator cancels out the remaining x^2 in the integral, leaving behind a $\frac{1}{3}$ . We can pull the $\frac{1}{3}$ out front of the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{x^2e^{x^3}dx}$

2. u=x^3

3. du=3x^2

4. $\int{x^2e^u(\frac{du}{3x^2})}$

5. $\frac{1}{3}\int{e^udu}$

6. $\frac{1}{3}e^u+c$

7. $\frac{1}{3}e^{x^3}+c$

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Example Question #6 : Use Substitution And Integration

Solve the following integral using substitution:

$\int{e^xcos(e^x)dx}$

Possible Answers:

-sin(e^x)+c

$e^{cosx}+c$

-cos(e^x)+c

sin(e^x)+c

$e^{sinx}+c$

Correct answer:

sin(e^x)+c

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression e^x . Next, we must take the derivative of . Its derivative is e^x . Next, solve this equation for so that we may replace it in the integral. Plug in place of e^x and $\frac{du}{e^x}$ in place of into the original integral and simplify. The e^x in the denominator cancels out the remaining e^x in the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{e^xcos(e^x)dx}$

2. u=e^x

3. du=e^xdx

4. $\int{e^xcosu(\frac{du}{e^x})}$

5. $\int{cosudu}$

6. sinu+c

7. sin(e^x)+c

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Example Question #7 : Use Substitution And Integration

Solve the following integral using substitution:

$\int{\frac{sin(lnx)}{x}dx}$

Possible Answers:

$\frac{1}{\sqrt{1-lnx}}+c$

$-sinx(\frac{1}{x})+c$

cos(lnx)+c

-cos(lnx)+c

$sinx(\frac{1}{x})+c$

Correct answer:

-cos(lnx)+c

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression lnx . Next, we must take the derivative of . Its derivative is $\frac{1}{x}$ . Next, solve this equation for so that we may replace it in the integral. Plug in place of lnx and dux in place of into the original integral and simplify. The in the denominator cancels out the remaining in the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{\frac{sin(lnx)}{x}dx}$

2. u=lnx

3. $du=(\frac{1}{x}dx)$

4. dux=dx

5. $\int{\frac{sinu}{x}dux}$

6. $\int{sinudu}$

7. -cosu+c

8. -cos(lnx)+c

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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 #1 : Complete The Square With Integration

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

Possible Answers:

To use the quadratic formula

Only when we are asked to in a problem

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

To find the roots of a function

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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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #1 : Use Substitution And Integration

Example Question #2 : Use Substitution And Integration

Example Question #3 : Use Substitution And Integration

Example Question #4 : Use Substitution And Integration

Example Question #5 : Use Substitution And Integration

Example Question #6 : Use Substitution And Integration

Example Question #7 : Use Substitution And Integration

Example Question #1 : Complete The Square With Integration

Example Question #1 : Complete The Square With Integration

Example Question #1 : Complete The Square With Integration

All Calculus AB Resources

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