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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 #54 : Indefinite Integrals

Evaluate.

$\int \frac{1}{2}x - \frac{1}{3} x^2 \ dx$

Possible Answers:

F(x) = x^2 - x^3

$F(x) = \frac{1}{4}x - \frac{1}{9} x^2$

$F(x) = \frac{1}{2}x^2 - \frac{1}{3}x^3$

Answer not listed.

$F(x) = \frac{1}{4}x^2 - \frac{1}{9} x^3$

Correct answer:

$F(x) = \frac{1}{4}x^2 - \frac{1}{9} x^3$

Explanation:

$\int f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \frac{1}{2}x - \frac{1}{3} x^2$ .

The antiderivative is $F(x) = \frac{1}{4}x^2 - \frac{1}{9} x^3$ .

Report an Error

Example Question #53 : Indefinite Integrals

Evaluate.

$\int \cos (2x-1) \ dx$

Possible Answers:

$F(x) = \frac{\cos (2x-1)}{x}$

Answer not listed.

$F(x) = 2 \sin (2x-1)$

$F(x) = 2 \cos (2x-1)$

$F(x) = \frac{\sin (2x-1)}{2}$

Correct answer:

$F(x) = \frac{\sin (2x-1)}{2}$

Explanation:

$\int f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \cos (2x-1)$ .

The antiderivative is $F(x) = \frac{\sin (2x-1)}{2}$ .

Report an Error

Example Question #56 : Indefinite Integrals

Evaluate.

$\int \frac{1}{4x-5} \ dx$

Possible Answers:

$F(x) = \frac{ \ln (x)}{4x-5}$

$F(x) = 4\ln (4x-5)$

Answer not listed.

$F(x) = \frac{ \ln (4x-5)}{4}$

$F(x) = \frac{ \ln (4x-5)}{\ln(x)}$

Correct answer:

$F(x) = \frac{ \ln (4x-5)}{4}$

Explanation:

$\int f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \frac{1}{4x-5}$ .

The antiderivative is $F(x) = \frac{ \ln (4x-5)}{4}$ .

Report an Error

Example Question #57 : Indefinite Integrals

Evaluate.

$\int 2x + 3x^2 + 4x^3 \ dx$

Possible Answers:

F(x) =2 + 6x + 12x^2

F(x) =2x^2 + 3x^3 + 4x^4

$F(x) =\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4}$

Answer not listed.

F(x) =x^2 + x^3 + x^4

Correct answer:

F(x) =x^2 + x^3 + x^4

Explanation:

$\int f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, f(x) = 2x + 3x^2 + 4x^3 .

The antiderivative is F(x) =x^2 + x^3 + x^4 .

Report an Error

Example Question #58 : Indefinite Integrals

Evaluate.

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

Possible Answers:

$F(x) = \frac{e^{x}}{3}$

$F(x) = \frac{e^{3x- 2}}{3}$

$F(x) = e^{3x- 2}$

Answer not listed.

$F(x) = e^{x}$

Correct answer:

$F(x) = \frac{e^{3x- 2}}{3}$

Explanation:

$\int f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = e^{3x- 2}$ .

The antiderivative is $F(x) = \frac{e^{3x- 2}}{3}$ .

Report an Error

Example Question #59 : Indefinite Integrals

Evaluate.

$\int 3x^2 + 12x^{11} \ dx$

Possible Answers:

Answer not listed

$F(x) = x^3 + x^{12}$

$F(x) = x^2 + x^{11}$

$F(x) = 3x^3 + 12x^{12}$

$F(x) = \frac{1}{3}x^3 +\frac{1}{12} x^{12}$

Correct answer:

$F(x) = x^3 + x^{12}$

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = 3x^2 + 12x^{11}$ .

The antiderivative is $F(x) = x^3 + x^{12}$ .

Report an Error

Example Question #54 : Indefinite Integrals

Evaluate.

$\int e^{4x- 2} \ dx$

Possible Answers:

Answer not listed

$F(x) = \frac{e^{x} }{4}$

$F(x) = \frac{e^{4x- 2} }{4}$

$F(x) = \frac{e^{x} }{4x-2}$

$F(x) = e^{4x- 2}$

Correct answer:

$F(x) = \frac{e^{4x- 2} }{4}$

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = e^{4x- 2}$ .

The antiderivative is $F(x) = \frac{e^{4x- 2} }{4}$ .

Report an Error

Example Question #681 : Integrals

Integrate:

$\int x^2e^xdx$

Possible Answers:

e^x(x^2+2x-2)+C

e^x(x^2-2x)+C

e^x(x^2-2x+2)+C

e^x(x^2-2x+2)

Correct answer:

e^x(x^2-2x+2)+C

Explanation:

To solve the integral, we must use integration by parts, which states that

$\int udv=uv-\int vdu$

We must designate a u and a dv:

u=x^2, du=2xdx

dv=e^xdx, v=e^x

The derivative and integral of u and dv, respectively, were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\int e^xdx=e^x+C$

Using the above formula, we can rewrite the integral:

$\int x^2e^xdx=x^2e^x-\int 2xe^xdx$

We must use integration by parts again, because we have an integral that cannot be solved using other methods.

This time, using the same integration and derivation rules, we have

u=2x, du=2dx

dv=e^xdx, v=e^x

Using the same formula above, we get

$\int x^2e^xdx=x^2e^x-\int 2xe^xdx=x^2e^x-[2xe^x-\int 2e^xdx]$

Integrating, we get

$x^2e^x-[2xe^x-\int 2e^xdx]=x^2e^x-2xe^x+2e^x+C$

The same integration rule above was used.

Rewriting, our final answer is

e^x(x^2-2x+2)+C

Report an Error

Example Question #332 : Finding Integrals

Evaluate.

$\int \cos (2x) - \frac{1}{2x} \ dx$

Possible Answers:

$F(x) = \frac{ \sin(2x)}{2} - \frac{\ln|2x|}{2}$

$F(x) = \frac{ \sin(x)}{2} - \frac{\ln(x)}{2}$

Answer not listed.

$F(x) = \sin(2x) - \ln(2x)$

$F(x) = \frac{ \sin(2x)}{4} + \frac{\ln(2x)}{2}$

Correct answer:

$F(x) = \frac{ \sin(2x)}{2} - \frac{\ln|2x|}{2}$

Explanation:

$\int f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \cos (2x) - \frac{1}{2x}$ .

The antiderivative is $F(x) = \frac{ \sin(2x)}{2} - \frac{\ln|2x|}{2}$ .

Report an Error

Example Question #682 : Integrals

Evaluate.

$\int \frac{1}{9x-8} \ dx$

Possible Answers:

Answer not listed.

$F(x) = \ln (9x-8)$

$F(x) = \frac{\ln (9x-8)}{x}$

$F(x) = \frac{9x-8}{\ln (9x-8)}$

$F(x) = \frac{\ln (9x-8)}{9}$

Correct answer:

$F(x) = \frac{\ln (9x-8)}{9}$

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \frac{1}{9x-8}$ .

The antiderivative is $F(x) = \frac{\ln (9x-8)}{9}$ .

Report an Error

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #54 : Indefinite Integrals

Example Question #53 : Indefinite Integrals

Example Question #56 : Indefinite Integrals

Example Question #57 : Indefinite Integrals

Example Question #58 : Indefinite Integrals

Example Question #59 : Indefinite Integrals

Example Question #54 : Indefinite Integrals

Example Question #681 : Integrals

Example Question #332 : Finding Integrals

Example Question #682 : Integrals

All Calculus 2 Resources

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