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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2371 : Calculus Ii

Give the solution of the differential equation

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

that satisfies the initial condition y(1)= 2 .

Possible Answers:

$y=2 e^ { x -1 }$

y =2 x^2

$y=2e^ { x^2 -1 }$

$y =2 + \ln x$

y = x^2+1

Correct answer:

$y=2e^ { x^2 -1 }$

Explanation:

As a separable differential equation, this can be solved as follows:

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

$\frac{\frac{\mathrm{d} y}{\mathrm{d} x} \cdot \mathrm{d} x }{y}= \frac{2xy \cdot \mathrm{d} x}{y}$

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

Now, integrate both sides:

$\int \frac{\mathrm{d} y }{y}=\int 2x \; \mathrm{d} x$

$\ln y = x^2 + C$

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

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

$y= e^ { C} \cdot e^ { x^2 }$

$y= C_{0} e^ { x^2 }$

Since the initial condition is y(1)= 2 , we can find $C_{0}$ by substituting:

$2= C_{0} e^ { 1^2 }$

$2= C_{0} e$

$\frac{2}{e}= \frac{C_{0} e}{e} = C_{0}$

The solution is:

$y=\frac{2}{e}\cdot e^ { x^2 }$

$y=2 \cdot \frac{e^ { x^2 }}{e}$

$y=2e^ { x^2 -1 }$

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Example Question #273 : Finding Integrals

Solve the differential equation

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

subject to the initial condition

$y (0)= \frac{\pi }{6}$ .

Possible Answers:

$y =\sin ^{-1} \left ( - \tan x + \frac{1}{2} \right )$

$y =\sin ^{-1} \left ( - \tan x + \frac{\sqrt{3}}{2} \right )$

$y =\sin ^{-1} \left ( - \tan x + \frac{\sqrt{2}}{2} \right )$

$y =\cos ^{-1} \left ( - \tan x + \frac{1}{2} \right )$

$y =\cos ^{-1} \left ( - \tan x + \frac{\sqrt{3}}{2} \right )$

Correct answer:

$y =\cos ^{-1} \left ( - \tan x + \frac{\sqrt{3}}{2} \right )$

Explanation:

As a separable differential equation,

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

can be rewritten and solved as follows:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \sec^2 x \cdot \frac{1}{\sin y }$

$\frac{\mathrm{d} y}{\mathrm{d} x} \cdot \sin y \cdot \mathrm{d} x= \sec^2 x \cdot \frac{1}{\sin y } \cdot \sin y \cdot \mathrm{d} x$

$\sin y \; \mathrm{d} y = \sec^2 x \; \mathrm{d} x$

$\int \sin y \; \mathrm{d} y = \int \sec^2 x \; \mathrm{d} x$

$-\cos y = \tan x + C$

$\cos y = - \tan x - C$

To find , use the initial condition $y (0)= \frac{\pi }{6}$ :

$\cos \frac{\pi }{6} = - \tan 0 - C$

$\frac{ \sqrt{3}}{2} = 0 - C$

$C = -\frac{ \sqrt{3}}{2}$ , so

$\cos y = - \tan x + \frac{\sqrt{3}}{2}$

and

$y =\cos ^{-1} \left ( - \tan x + \frac{\sqrt{3}}{2} \right )$

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Example Question #274 : Finding Integrals

Give the solution of the differential equation

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

that satisfies the initial condition

y (0)= 10 .

Possible Answers:

$y= 10e^{ e^{x}- 1}$

$y= e^{ e^{x}-1} + 9$

$y = e^{x+1}+10 - e$

$y= 10e^{ e^{x} } - 10e + 10$

$y = e^{x}+9$

Correct answer:

$y= 10e^{ e^{x}- 1}$

Explanation:

As a separable differential equation, this can be solved as follows:

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

$\frac{\mathrm{d} y}{\mathrm{d} x} \cdot \frac{\mathrm{d} x}{y} = ye^{x} \cdot \frac{\mathrm{d} x}{y}$

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

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

$\ln y = e^{x} + C$

$e^ {\ln y} = e^{ e^{x}+ C}$

$y =e^{ C} \cdot e^{ e^{x}}$

$y =C_{0}\cdot e^{ e^{x}}$

To find $C_{0}$ , substitute the initial conditions:

$10=C_{0}\cdot e^{ e^{0}}$

$10=C_{0}\cdot e^{ 1}$

$10=C_{0}\cdot e$

$\frac{10}{e}=C_{0}$

The solution is $y =\frac{10}{e}\cdot e^{ e^{x}} = \frac{10e^{ e^{x}}} {e} = 10e^{ e^{x}- 1}$ .

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Example Question #275 : Finding Integrals

Give the solution of the differential equation

$\frac{\mathrm{d} y}{\mathrm{d} x} = \sec x \csc y$

that satisfies the initial condition

$y (0) = - \frac{\pi }{2}$ .

Possible Answers:

$y = \cos^{-1} \left (- \ln | \sec x + \tan x + e | \right )$

$y = \cos^{-1} \left ( \ln | \sec x - \tan x | \right )$

$y = \sin^{-1} \left ( \ln | \sec x - \tan x | \right )$

$y = \cos^{-1} \left (- \ln | \sec x + \tan x | \right )$

$y = \sin^{-1} \left ( \ln | \sec x - \tan x + e | \right )$

Correct answer:

$y = \cos^{-1} \left (- \ln | \sec x + \tan x | \right )$

Explanation:

As a separable differential equation, this can be solved as follows:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \sec x \csc y$

$\frac{\mathrm{d} y}{\mathrm{d} x} \cdot\frac{ \mathrm{d} x }{ \csc y}= \sec x \csc y \cdot\frac{ \mathrm{d} x }{ \csc y}$

$\frac{ \mathrm{d} y }{ \csc y}= \sec x \; \mathrm{d} x$

$\sin y \; \mathrm{d} y = \sec x \; \mathrm{d} x$

$\int \sin y \; \mathrm{d} y = \int \sec x \; \mathrm{d} x$

$-\cos y = \ln | \sec x + \tan x | + C$

$\cos y = - \ln | \sec x + \tan x | + C$

$y = \cos^{-1} \left (- \ln | \sec x + \tan x | + C \right )$

Apply the initial condition to find :

$-\frac{\pi}{2} = \cos^{-1} \left (- \ln | \sec 0 + \tan 0| + C \right )$

$-\frac{\pi}{2} = \cos^{-1} \left (- \ln |1 +0 + C | \right )$

$\cos \left (-\frac{\pi}{2} \right )= - \ln |1 +0 + C |$

$0= - \ln |C +1 |$

$0= \ln |C +1 |$

$e^0=e ^ { \ln |C +1 |}$

1 = C + 1

C= 0

The solution is

$y = \cos^{-1} \left (- \ln | \sec x + \tan x | \right )$ .

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Example Question #2372 : Calculus Ii

Solve the differential equation

$\frac{\mathrm{d} y}{\mathrm{d} x} = xy + y$

subject to the initial condition

y (0) = 3 .

Possible Answers:

$y=3 e^ { x^2 + x }$

$y=3 e^ { x^2 }$

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

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

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

Correct answer:

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

Explanation:

As a separable differential equation,

$\frac{\mathrm{d} y}{\mathrm{d} x} = xy + y$

can be rewritten as follows:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \left (x+ 1 \right ) y$

$\frac{\mathrm{d} y}{\mathrm{d} x} \cdot \frac{\mathrm{d} x}{y} = \left (x+ 1 \right ) y \cdot \frac{\mathrm{d} x}{y}$

$\frac{\mathrm{d} y}{y} = \left (x+ 1 \right ) \mathrm{d} x$

$\int \frac{\mathrm{d} y}{y} = \int \left (x+ 1 \right ) \mathrm{d} x$

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

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

$y=C_{0} e^ { \frac{x^2}{2} + x }$

We can find $C_{0}$ using the initial condition y (0) = 3 :

$3=C_{0} e^ { \frac{0^2}{2} + 0 } = C_{0} e^ { 0 } = C_{0} \cdot 1 = C_{0}$

The solution is

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

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Example Question #277 : Finding Integrals

Determine the indefinite integral:

$\int xe^{3x} \;\mathrm{d}x$

Possible Answers:

$\frac{3xe^{3x}-e^{3x}}{9} + C$

$\frac{ xe^{3x}-3e^{3x}}{9} + C$

$\frac{ xe^{3x}}{9} + C$

$\frac{3xe^{3x}-e^{3x}}{3} + C$

$\frac{ xe^{3x}-e^{3x}}{3} + C$

Correct answer:

$\frac{3xe^{3x}-e^{3x}}{9} + C$

Explanation:

Integration by parts is the best strategy here.

Let u = x and $\mathrm {d}v = e^{3x} \; \mathrm {d}x$ .

Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = 1$ and $\mathrm{d} u = \mathrm{d} x$ .

Also,

$\mathrm {d}v = e^{3x} \; \mathrm {d}x$ .

$\int \mathrm {d}v = \int e^{3x} \; \mathrm {d}x$

$v = \frac{e^{3x}}{3} = \frac{1}{3}e^{3x}$

Therefore,

$\int xe^{3x} \;\mathrm{d}x$

$= \int u \; \mathrm{d}v$

$= uv- \int v \; \mathrm{d}u$

$= x \cdot \frac{1}{3}e^{3x}- \int \frac{1}{3}e^{3x} \cdot \; \mathrm{d}x$

$= \frac{1}{3}xe^{3x}- \frac{1}{3} \int e^{3x} \mathrm{d}x$

$= \frac{1}{3}xe^{3x}- \frac{1}{3}\cdot \left ( \frac{e^{3x}}{3} + C \right )$

$= \frac{1}{3}xe^{3x}- \frac{1}{9} e^{3x} + C$

$= \frac{3xe^{3x}-e^{3x}}{9} + C$

Note the absorption of the negative into the constant in the third to last step.

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Example Question #278 : Finding Integrals

Solve the differential equation

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

subject to the initial condition

y(0) = e^2 .

Possible Answers:

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

$y = e^ { e^{x^2}+1 }$

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

$y = e^{x^2+1}+e^2 - e$

$y = e^ { e^{x^2+2x } +1}$

Correct answer:

$y = e^ { e^{x^2}+1 }$

Explanation:

As a separable differential equation,

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

can be rewritten and solved as follows:

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

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

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

The integral on the right can be solved by setting u = x^2 and, subsequently, $\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \; x^2 = 2x$ and $\mathrm{d} u = 2x \; \mathrm{d} x$ .

$\int \frac{\mathrm{d} y}{y} = \int e^{u}\mathrm{d} u$

$\ln y = e^u + C$

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

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

$y = C_{0} e^ { e^{x^2} }$

To find $C_{0}$ , use the initial conditions:

$e^2= C_{0} e^ { e^{0^2} }$

$e^2= C_{0} e^ { e^{0} }$

$e^2= C_{0} e^ { 1}$

$e^2= C_{0} e$

$C_{0} = \frac{ e^2 }{e} = e$

The solution is

$y = e \cdot e^ { e^{x^2} } = e^ { e^{x^2}+1 }$ .

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Example Question #1 : Indefinite Integrals

Find the indefinite integral of the following function:

$f(x)=\frac{2x}{x^2+1}dx$

Possible Answers:

ln|x^2+1|+C

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

ln(x^2+1)

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

$\frac{ln(x^2+1)}{(x^2+1)^2}+C$

Correct answer:

ln|x^2+1|+C

Explanation:

To integrate this function, use u substitution. Make,

$\\u=x^2+1 \\du=2xdx$

then substitute them into the equation to get

$f(x)=\frac{du}{u}$ .

The integral of

$\frac{du}{u}=ln|u|+C$

then plug u back into the equation

ln|x^2+1|+C .

The +C is essential because the integral is indefinite.

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Example Question #1 : Indefinite Integrals

Evaluate the given indefinite integral

$\int sec^2(x)tan(x)dx$ .

Possible Answers:

$\frac{sec^2(x)}{2}+C$

-ln(tan(x))+C

$\frac{sec^2(x)}{2}+C$

$\frac{sec^2(x)}{2}$

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

Correct answer:

$\frac{sec^2(x)}{2}+C$

Explanation:

To integrate this function, use u substitution. Make

$\\u=sec(x) \\du=sec(x)tan(x)dx$

then substitute them into the equation to get

$\int udu$ .

The integral of

$udu=\frac{u^2}{2}+C$

then plug u back into the equation

$\frac{sec^2(x)}{2}+C$ .

The +C is essential because the integral is indefinite.

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Example Question #3 : Indefinite Integrals

Evaluate the given indefinite integral

$\int -\frac{x}{\sqrt{1-x^2}}$ .

Possible Answers:

${1-x^2}$

$\sqrt{1-x^2}$

$\sqrt{1-x^2}+C$

$\frac{1}{\sqrt{1-x^2}}+C$

$\frac{1}{1-x^2}+C$

Correct answer:

$\sqrt{1-x^2}+C$

Explanation:

To integrate this function, use u substitution. Make

$\\u=1-x^2 \\du=-2xdx$

then substitute them into the equation to get

$\int \frac{\frac{1}{2}du}{\sqrt{u}}$ .

The integral of

$\int \frac{du}{\sqrt{u}}=2\sqrt{u}+C$

so we have

$\frac{1}{2}\cdot2\sqrt{u}+C=\sqrt{u}+C=\sqrt{1-x^2}+C$ .

The +C is essential because the integral is indefinite.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2371 : Calculus Ii

Example Question #273 : Finding Integrals

Example Question #274 : Finding Integrals

Example Question #275 : Finding Integrals

Example Question #2372 : Calculus Ii

Example Question #277 : Finding Integrals

Example Question #278 : Finding Integrals

Example Question #1 : Indefinite Integrals

Example Question #1 : Indefinite Integrals

Example Question #3 : Indefinite Integrals

All Calculus 2 Resources

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