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AP Calculus AB : Applications of antidifferentiation

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Example Questions

Example Question #31 : Integrals

If f(x)=x^2+2x+4 what is f'(x) ?

Possible Answers:

2(x+2)

2x+4

2x+2

Correct answer:

2x+2

Explanation:

Taking the derivative of x^2 gives you .

Taking the derivative of gives you .

Finally taking the derivative of gives you .

Therefore f'(x)=2x+2 .

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Example Question #9 : Solving Separable Differential Equations And Using Them In Modeling

Solve the following separable differential equation y' = ye^x with initial condition y(0) = e^2 .

Possible Answers:

$y=e^{e^x}$

$y=ee^{e^x}$

$y=e^{x^2}$

y=e^2e^x

$y=e^2e^{e^x}$

Correct answer:

$y=ee^{e^x}$

Explanation:

We proceed as follows

y'=ye^x . Start

$\frac{dy}{dx} = ye^x$ . Rewrite as $\frac{dy}{dx}$ .

$\frac{1}{y}dy=e^xdx$ . Multiply both sides by , and divide both sides by .

$\ln|y| =e^x +C$ . Integrate both sides. Do not forget the on one of the sides.

Substitute the initial condition y(0)=e^2 .

$\ln|e^2| = e^0+C$ .

1=C . Solve for .

$\ln|y| = e^x+1$

$y = e^{e^x+1}$ . Exponentiate both sides .

$y = ee^{e^x}$ . Rule of exponents.

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Example Question #10 : Solving Separable Differential Equations And Using Them In Modeling

Solve the separable, first-order differential equation for :

$(y+1)\frac{dy}{dx}=2x-8+\frac{dy}{dx}$

y(0)=8

Possible Answers:

$y =-\sqrt{2x^2 -16x+8 }$

y = x^2 - 8x+8

$y =\sqrt{2x^2 -16x+16 }$

$y =\sqrt{x^2 -4x+8 }$

y = 8x-x^2 -8

Correct answer:

$y =\sqrt{2x^2 -16x+16 }$

Explanation:

Solve the separable, first-order differential equation for :

$(y+1)\frac{dy}{dx}=2x-8+\frac{dy}{dx}$

First collect all the terms with the derivative to one side of the equation.

$(y+1)\frac{dy}{dx}-\frac{dy}{dx}=2x-8$

$\frac{dy}{dx}[(y+1)-1]=2x-8$

$y\frac{dy}{dx}=2x-8$

Important Conceptual Note: often in texts on differential equations differentials often appear to have been rearranged algebraically as if $\frac{dy}{dx}$ is a "fraction," making it appear as if we "multiplied both sides" by to get: . This is not the case. The derivative is a limit by definition and, when the limit exists, can take on any real number which includes irrational numbers i.e. numbers which cannot be written as a ratio of two integers.

For instance, we cannot represent $\pi$ as a ratio, but some functions may have a derivative at a point such that the derivative is equal to $\pi$ , or a funciton may simply have an irrational number like $\pi$ as a derivative. For instance, if $y=\pi x$ we write the derivative $\frac{dy}{dx}=\pi$ . Claiming that and are representative of a "numerator" and "denominator" respectively, we would essentially be claiming to have found a way to write an irrational number, such as $\pi$ as a ratio, which is preposterous. The expression $\frac{dy}{dx}$ is simply notation.

Here is what we are really doing.

$\int y\frac{dy}{dx}dx=\int (2x-8)dx$

$\int y dy=\int (2x-8)dx$

$\frac{1}{2}y^2 +C_1 =x^2 -8x +C_2$

Note that the constants of integration can just be combined into one constant by defining .

Solve for :

y^2 =2(x^2 -8x +C)

$y =\pm \sqrt{2x^2 -16x+C }$

Applying the initial condition:

$y(0)=\pm \sqrt{C}=8$

C = 16

$y =\pm \sqrt{2x^2 -16x+16 }$

Here we have two possible solutions. However, because of the initial condition, we can easily rule out the negative solution. y(0) must be equal to positive .

$y(0)=8\: \: \: \: \: \: \Rightarrow\: \: \: \: \: \:\sqrt{16}=8$

$y =\sqrt{2x^2 -16x+16 }$

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Example Question #11 : Applications Of Antidifferentiation

Solve the separable differential equation

$\frac{\mathrm{d} y}{\mathrm{d} x}=x^2y$

given the condition y(0)=3

Possible Answers:

y=0

None of the other answers

$y=Ce^{\frac{x^3}{3}}$

y=C

$y=3e^{\frac{x^3}{3}}$

Correct answer:

$y=3e^{\frac{x^3}{3}}$

Explanation:

To solve this equation, we must separate the variables such that terms containing x and y are on the same side as dx and dy, respectively:

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

Integrating both sides of the equation, we get

$\ln \left | y\right |+C_1=\frac{x^3}{3}+C_2$

The integrals were found using the following rules:

$\int \frac{dx}{x}=\ln \left | x\right |+C$ , $\int x^n dx=\frac{x^{n+1}}{n+1}+C$

After combining the constants of integration into a single C, exponentiating both sides, and using the properties of exponents to simplify, we get

$y=e^{\frac{x^3}{3}+C}=e^{\frac{x^3}{3}}\cdot e^C=Ce^{\frac{x^3}{3}}$

To solve for C, we use the condition given:

3=Ce^0

3=C

Our final answer is

$y=3e^{\frac{x^3}{3}}$

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Example Question #12 : Applications Of Antidifferentiation

is a function of . Solve for in this differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \sqrt{\frac{ e^{y}}{x}}$

Possible Answers:

$y =-2 \ln ( \sqrt{x}+ C)$

$y =- \ln (- \sqrt{x}+ C)$

$y =\ln ( \sqrt{x}+ C)$

$y =-2 \ln (- \sqrt{x}+ C)$

$y =2 \ln (- \sqrt{x}+ C)$

Correct answer:

$y =-2 \ln (- \sqrt{x}+ C)$

Explanation:

First, rewrite the expression on the right as the $\frac{1}{2}$ power of the radicand:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \sqrt{\frac{ e^{y}}{x}}$

$\frac{\mathrm{d} y}{\mathrm{d} x} =\left ( \frac{ e^{y}}{x} \right )^{\frac{1}{2}}$

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{ (e^{y})^{\frac{1}{2}}}{x^{\frac{1}{2}}}$

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{ e^ {\frac{y}{2}}}{x^{\frac{1}{2}}}$

The expressions with can be separated from those with by multiplying both sides by $e^ {-\frac{y}{2} }\mathrm{\; d} x$ :

$\frac{\mathrm{d} y}{\mathrm{d} x} \cdot e^ {-\frac{y}{2} }\mathrm{\; d} x = \frac{ e^ {\frac{y}{2}}}{x^{\frac{1}{2}}} \cdot e^ {-\frac{y}{2} }\mathrm{\; d} x$

$e^ {-\frac{y}{2} } \mathrm{d} y= \frac{ 1}{x^{\frac{1}{2}}} \mathrm{\; d} x$

$e^ {-\frac{1}{2} y} \mathrm{d} y= x^{-\frac{1}{2}} \mathrm{\; d} x$

Find the indefinite integral of both sides:

$\int e^ {-\frac{1}{2} y} \mathrm{d} y= \int x^{-\frac{1}{2}} \mathrm{\; d} x$

The expression on the right can be integrated using the Power Rule. On the right, use some -substitution, setting $u = -\frac{1}{2} y$ ; this makes $\mathrm{d} u = -\frac{1}{2} \mathrm{d}y$ and $\mathrm{d}y = -2 \mathrm{\; d} u$ :

$-2 \int e^ {u} \mathrm{d} u= \frac{x\frac{1}{2}}{\frac{1}{2}}+ C$

$-2 e^ {u} =2 \sqrt{x}+ C$

Apply some algebra to solve for :

$e^ {u} =- \sqrt{x}+ C$

$u =\ln (- \sqrt{x}+ C)$

Substitute $-\frac{1}{2} y$ back for , and apply some algebra:

$-\frac{1}{2} y =\ln (- \sqrt{x}+ C)$

$y =-2 \ln (- \sqrt{x}+ C)$

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Example Question #13 : Solving Separable Differential Equations And Using Them In Modeling

Solve the separable differential equation

$\frac{\mathrm{d} y}{\mathrm{d} x}=y(x^2+5x+2)$

where y(0)=7

Possible Answers:

$y=7e^{\frac{x^3}{3}+\frac{5x^2}{2}+2x}$

y=0

$y=e^{\frac{x^3}{3}+\frac{5x^2}{2}+2x}$

y=7

Correct answer:

$y=7e^{\frac{x^3}{3}+\frac{5x^2}{2}+2x}$

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

$\frac{dy}{y}=(x^2+5x+2)dx$

Next, we integrate both sides:

$\ln \left | y\right |=\frac{x^3}{3}+\frac{5x^2}{2}+2x+C$

The integrals were solved using the following rules:

$\int \frac{dx}{x}=\ln \left | x\right |+C$ , $\int x^n dx=\frac{x^{n+1}}{n+1}+C$

The two constants of integration were combined to make a single one.

Now, we exponentiate both sides to solve for y, keeping in mind rules for exponents which allow us to move the integration constant to the front:

$y=Ce^{\frac{x^3}{3}+\frac{5x^2}{2}+2x}$

To solve for the constant of integration, we use the condition given:

$7=Ce^{0}=C(1)=C$

Our final answer is

$y=7e^{\frac{x^3}{3}+\frac{5x^2}{2}+2x}$

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Example Question #14 : Solving Separable Differential Equations And Using Them In Modeling

Solve the following separable differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x}= y\sec^2(x)$

given the condition that at $x=\pi, y=-2$

Possible Answers:

$y=Ce^{\tan(x)}$

$y=-2e^{\tan(x)}$

y=0

$y=2e^{\tan(x)}$

y=-2

Correct answer:

$y=-2e^{\tan(x)}$

Explanation:

To solve the separable differential equation, we must separate x and y and their respective derivatives to either side of the equal sign:

$\frac{dy}{y}=\sec^2(x)dx$

Now, we integrate both sides of the equation:

$\ln \left | y\right |=\tan(x)+C$

The integrals were found using their identical rules.

Exponentiating both sides of the equation to solve for y - and keeping in mind the rules of exponents - we get

$y=e^{\tan(x)+C}=e^C\cdot e^{\tan(x)}=Ce^{\tan(x)}$

Now, we solve for the integration constant by using the condition given:

$-2=Ce^{\tan(\pi)}=Ce^0=C$

Our final answer is

$y=-2e^{\tan(x)}$

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Example Question #11 : Applications Of Antidifferentiation

Solve the following separable differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{x^2}{y\cos(y^2)}$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

$y\cos(y^2)dy=x^2dx$

Next, we integrate both sides, where on the lefthand side, the following substitution is made:

u=y^2, du=2ydy

$\frac{1}{2}\int\cos(u)du=\frac{1}{2}\sin(y^2)=\frac{x^3}{3}+C$

The integrals were solved using the following rules:

$\int \cos(x)dx=\sin(x)+C, \int x^ndx=\frac{x^{n+1}}{n+1}+C$

The two constants of integration were combined to make a single one.

Now, we solve for y:

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

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

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Example Question #16 : Solving Separable Differential Equations And Using Them In Modeling

Solve the separable differential equation

$\frac{\mathrm{d} y}{\mathrm{d} x} = y(2x+x^2)$

given the initial condition

f(0)=14

Possible Answers:

$y=14e^{\frac{x^2}{2}+\frac{x^3}{3}}$

$y=14e^{2x+\frac{x^3}{3}}$

$y=14e^{x^2+\frac{x^2}{3}}$

$y=14e^{x^2+\frac{x^3}{3}}$

$y=Ce^{x^2+\frac{x^3}{3}}$

Correct answer:

$y=14e^{x^2+\frac{x^3}{3}}$

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

$\frac{dy}{y} = (2x+x^2)dx$

Next, we integrate both sides:

$\ln \left | y\right |= x^2+\frac{x^3}{3}+C$

The integrals were solved using the following rules:

$\int \frac{dx}{x}=\ln \left | x\right |+C$ , $\int x^n dx = \frac{x^{n+1}}{n+1}+C$

The two constants of integration were combined to make a single constant.

Now, exponentiate both sides to isolate y, and use the properties of exponents to rearrange the integration constant:

$e^{\ln\left | y\right |}= e^{x^2+\frac{x^3}{3}+C}$

$y=e^{x^2+\frac{x^3}{3}}\cdot e^C=Ce^{x^2+\frac{x^3}{3}}$

(The exponential of the constant is another constant.)

Finally, we solve for the integration constant using the initial condition:

14=Ce^0=C

Our final answer is

$y=14e^{x^2+\frac{x^3}{3}}$

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Example Question #16 : Solving Separable Differential Equations And Using Them In Modeling

Solve the separable differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{x^4+e^x}{2y}, y< 0$

Possible Answers:

$y=\sqrt{\frac{x^5}{5}+e^x}+C$

$y=-\sqrt{\frac{x^5}{5}+e^x+C}$

$y=-\sqrt{{x^5}+e^x+C}$

$y=\sqrt{\frac{x^5}{5}+e^x+C}$

Correct answer:

$y=-\sqrt{\frac{x^5}{5}+e^x+C}$

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

2y dy=(x^4+x)dx

Next, we integrate both sides:

$y^2=\frac{x^5}{5}+e^x+C$

The integrals were solved using the following rules:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ , $\int e^x dx=e^x+C$

The two constants of integration were combined to make a single one.

Now, we solve for y:

$y=\pm \sqrt{\frac{x^5}{5}+e^x+C}$

Because the problem statement said that y is negative - and y cannot be zero - our final answer is

$y=-\sqrt{\frac{x^5}{5}+e^x+C}$

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AP Calculus AB : Applications of antidifferentiation

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #31 : Integrals

Example Question #9 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #10 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #11 : Applications Of Antidifferentiation

Example Question #12 : Applications Of Antidifferentiation

Example Question #13 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #14 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #11 : Applications Of Antidifferentiation

Example Question #16 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #16 : Solving Separable Differential Equations And Using Them In Modeling

All AP Calculus AB Resources

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