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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 #42 : Indefinite Integrals

Evaluate the following indefinite integral:

$\int(9e^2^x+5cosx) dx$

Possible Answers:

18e^2^x+5sinx+C

18e^2^x-5sinx+C

4.5e^2^x+5sinx+C

4.5e^2x-5sinx+C

Correct answer:

4.5e^2^x+5sinx+C

Explanation:

The antiderivative of a sum is the sum of the antiderivatives. The first term in the integrand can be solved by u-substitution, as follows:

$\int9e^2^x dx = 9\int e^2^x dx = 9\int (e^u)(\frac{du}{2}) = 4.5e^u +C = 4.5e^2^x + C$

The second term in the integrand is a straightforward integration of a trigonometric function:

$\int5cosx dx = 5 \int cosx dx = 5sinx +C$

Hence, the antiderivative of the original function is

$\int(9e^2^x+5cosx)dx = (4.5e^2^x+C) +(5sinx + C)$

=4.5e^2^x+5sinx+C

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Example Question #42 : Indefinite Integrals

Evaluate.

$\int 2x + \frac{1}{x} \ dx$

Possible Answers:

$F(x) = x^2 + \ln(x)$

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

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

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

Answer not listed

Correct answer:

$F(x) = x^2 + \ln(x)$

Explanation:

$\int_{a}^{b} f(x) dx$

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

If then
If then $F(x) = \frac{x^{a+1}}{a+1} + C$
If $f(x) = \frac{1}{x}$ then
If then
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 $F(x) = \tan(x) + C$
Substitution rule $\int f (g(x)) g'(x) \ dx = \int f(u) \ du$ where

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

The antiderivative is $F(x) = x^2 + \ln(x)$ .

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Example Question #43 : Indefinite Integrals

Evaluate.

$\int \cos (2x-3) \ dx$

Possible Answers:

$F(x) = - \frac{\cos(2x-3)}{\sin(2x-3)}$

Answer not listed

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

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

$F(x) =- \sin(2x-3)$

Correct answer:

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

Explanation:

$\int_{a}^{b} f(x) dx$

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

If then
If then $F(x) = \frac{x^{a+1}}{a+1} + C$
If $f(x) = \frac{1}{x}$ then
If then
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 $F(x) = \tan(x) + C$
Substitution rule $\int f (g(x)) g'(x) \ dx = \int f(u) \ du$ where

In this case, $f(x) = \cos(2x-3)$ where u= 2x-3 .

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

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Example Question #41 : Indefinite Integrals

$\int cot(3x+3)dx$

Possible Answers:

ln|sin(3x+3)|+C

$\frac{1}{3}ln|sin(3x+3)|+C$

$\frac{1}{3}ln|cos(3x+3)|+C$

3ln|cos(3x+3)|+C

3ln|sin(3x+3)|+C

Correct answer:

$\frac{1}{3}ln|sin(3x+3)|+C$

Explanation:

This is a u-substitution problem. First, let us rewrite our trigonometric function:

$\int cot(3x+3)dx=\int \frac{cos(3x+3)}{sin(3x+3)}dx.$

We want to replace our variables to make the integral easier to solve.

u=sin(3x+3), du=3cos(3x+3)dx.

Now, we have replaced everything in our integral in terms of our new variable However, we have an extra that was not part of our original function. Therefore, we must divide by so that we match everything exactly. Rewriting, we get:

$\frac{1}{3}\int \frac{du}{u}=\frac{1}{3}ln|u|+C.$ Now, all we have to is substitute our original variable back in.

$\frac{1}{3}ln|sin(3x+3)|+C.$

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Example Question #322 : Finding Integrals

Evaluate.

$\int 12x^5 - 3x^2 \ dx$

Possible Answers:

F(x) =12x^3 - 3x^4

Answer not listed

F(x) =x^6 - x^3

F(x) =6x

F(x) =2x^6 - x^3

Correct answer:

F(x) =2x^6 - 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) = 12x^5 - 3x^2 .

The antiderivative is F(x) =2x^6 - x^3 .

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Example Question #675 : Integrals

Evaluate.

$\int \cos (3x) - 1\ dx$

Possible Answers:

$F(x) = - \frac{\sin(3x)}{3x} - x$

$F(x) = \frac{\sin(3x)}{3}$

$F(x) = \frac{\sin(3x)}{3} - x$

$F(x) = \frac{\sin(x)}{3}$

Answer not listed

Correct answer:

$F(x) = \frac{\sin(3x)}{3} - x$

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, $f(x) = \cos(3x) -1$ where u= 3x .

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

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Example Question #676 : Integrals

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

Possible Answers:

$F(x) =\frac{\ln(x)}{3}$

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

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

Answer not listed

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

Correct answer:

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

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, $f(x) = \frac{1}{3x-1}$ and u= 3x-1 .

The antiderivative is $F(x) =\frac{\ln(3x-1)}{3}$ .

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Example Question #51 : Indefinite Integrals

Evaluate.

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

Possible Answers:

Answer not listed

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

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

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

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

Correct answer:

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

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, $f(x) = e^{2x-4}$ and u= 2x-4 .

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

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Example Question #52 : Indefinite Integrals

Evaluate.

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

Possible Answers:

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

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

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

Answer not listed

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

Correct answer:

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

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) = (x-1)^2 .

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

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Example Question #331 : Finding Integrals

Evaluate.

$\int 6x^2 + \sin(x) \ dx$

Possible Answers:

$F(x) = 2x^3 - \cos(x)$

$F(x) = 2x^3 + \cos(x)$

Answer not listed

$F(x) = x^4 - \sin(x)$

$F(x) = x^4 - \cos(x)$

Correct answer:

$F(x) = 2x^3 - \cos(x)$

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, $f(x) = 6x^2 + \sin(x)$ .

The antiderivative is $F(x) = 2x^3 - \cos(x)$ .

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #42 : Indefinite Integrals

Example Question #42 : Indefinite Integrals

Example Question #43 : Indefinite Integrals

Example Question #41 : Indefinite Integrals

Example Question #322 : Finding Integrals

Example Question #675 : Integrals

Example Question #676 : Integrals

Example Question #51 : Indefinite Integrals

Example Question #52 : Indefinite Integrals

Example Question #331 : Finding Integrals

All Calculus 2 Resources

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