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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 #108 : Definite Integrals

Evaluate.

$\int_{1}^{2} 6x +4 \ dx$

Possible Answers:

Answer not listed.

Correct answer:

Explanation:

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

In this case, f(x) = 6x +4 .

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

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{1}^{2} 6x +4 \ dx = \left ( 3x^2 + 4x \right )_{1}^{2}$

$= 3(2)^2 + 4(2) - \left ( 3(1)^2 + 4(1) \right )$

=13

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Example Question #109 : Definite Integrals

Evaluate.

$\int_{3}^{4} \frac{1}{3x-7} \ dx$

Possible Answers:

Answer not listed.

0.31

1.54

0.14

3.69

Correct answer:

0.31

Explanation:

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

In this case, $f(x) = \frac{1}{3x-7}$ .

The antiderivative is $F(x) =\frac{\ln(3x-7)}{3}$ .

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{3}^{4} \frac{1}{3x-7} \ dx= \left ( \frac{\ln(3x-7)}{3} \right )_{3}^{4}$

$= \frac{\ln(3(4)-7)}{3} - \left ( \frac{\ln(3(3)-7)}{3} \right )$

=0.31

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Example Question #110 : Definite Integrals

Evaluate.

$\int_{0}^{2} x^6 \ dx$

Possible Answers:

$\frac{1}{3}$

$\frac{8}{3}$

$\frac{128}{7}$

$\frac{22}{3}$

Answer not listed.

Correct answer:

$\frac{128}{7}$

Explanation:

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

In this case, f(x) = x^6 .

The antiderivative is $F(x) = \frac{x^7}{7}$ .

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{0}^{2} x^6 \ dx = \left ( \frac{x^7}{7} \right )_{0}^{2}$

$= \frac{2^7}{7} - \left ( \frac{0^5}{7} \right )$

$= \frac{128}{7}$

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Example Question #111 : Definite Integrals

Evaluate.

$\int_{1}^{2} e^{43x -30} \ dx$

Possible Answers:

$\frac{e^{56}}{43}-\frac{e^{13}}{43}$

-0.75

Answer not listed.

5.21

6.54

Correct answer:

$\frac{e^{56}}{43}-\frac{e^{13}}{43}$

Explanation:

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

In this case, $f(x) = e^{43x -30}$ .

The antiderivative is $F(x) =\frac{ e^{43x -30}}{43}$ .

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{1}^{2} e^{43x -30} \ dx = \left ( \frac{ e^{43x -30}}{43} \right )_{1}^{2}$

$=\frac{ e^{43(2) -30}}{43} - \left ( \frac{ e^{43(1) -30}}{43} \right )$

$=\frac{e^{56}}{43}-\frac{e^{13}}{43}$

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Example Question #112 : Definite Integrals

$\int_{2}^{5}x^2+4x-1dx$

Possible Answers:

Correct answer:

Explanation:

First, integrate this expression. Remember to add one to the exponent and then put that result on the denominator:

$\frac{x^3}{3}+\frac{4x^2}{2}-x=\frac{x^3}{3}+2x^2-x$ .

Then, evaluate first at 5 and then at 2. Subtract those two results:

$(\frac{125}{3}+50-5)-(\frac{8}{3}+8-2)=\frac{117}{3}+39=78$ .

Thus, your answer is 78.

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Example Question #113 : Definite Integrals

$\int_{1}^{3}\frac{5+3x}{x}dx$

Possible Answers:

11.4978

11.4931

11.5731

12.4931

11.4631

Correct answer:

11.4931

Explanation:

First, chop up the fraction into two separate terms:

$\int_{1}^{3}\frac{5}{x}+3dx$

Now, integrate. Remember that when there is a single x on the denominator, the integral is ln of that term.

$5ln\left | x \right |+3x$

Evaluate at 3 and then at 1. Subtract the results:

(5ln3+9)-(5ln1+3)=11.4931 .

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Example Question #114 : Definite Integrals

$\int_{2}^{4}(x-1)(2x+3)dx$

Possible Answers:

$\frac{112}{5}$

$\frac{112}{3}$

$\frac{113}{3}$

$\frac{101}{3}$

$\frac{92}{3}$

Correct answer:

$\frac{112}{3}$

Explanation:

First, use FOIL to multiply the binomials before integrating:

$\int_{2}^{4}2x^2+x-3dx$

Then, integrate. Remember to add one to the exponent and then put that result on the denominator:

$\frac{2x^3}{3}+\frac{x^2}{2}-3x$

Evaluate at 4 and then 2. Subtract the results:

$(\frac{128}{3}+\frac{16}{2}-12)-(\frac{16}{3}+\frac{4}{2}-6)=\frac{112}{3}$

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Example Question #115 : Definite Integrals

Evaluate.

$\int_{1}^{2}12x-9 dx$

Possible Answers:

Answer not listed.

Correct answer:

Explanation:

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

In this case, f(x) = 12x-9 .

The antiderivative is F(x) = 6x^2-9x .

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{1}^{2}12x-9 dx = \left ( 6x^2-9x \right )_{1}^{2}$

$= 6(2)^2-9(2)- \left ( 6(1)^2-9(1) \right )$

= 8

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Example Question #116 : Definite Integrals

Evaluate.

$\int_{1}^{2} 13 \ dx$

Possible Answers:

Answer not listed

Correct answer:

Answer not listed

Explanation:

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

In this case, f(x) = 13 .

The antiderivative is F(x) = 13x .

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{1}^{2} 13 \ dx = \left ( 13x \right )_{1}^{2}$

$= 13(2) - \left ( 13(1) \right )$

= 13

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Example Question #117 : Definite Integrals

Evaluate.

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

Possible Answers:

0.46

1.08

2.11

Answer not listed.

-1.62

Correct answer:

1.08

Explanation:

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

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

The antiderivative is $F(x) = -3 \ln(3-x) - x$ .

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{1}^{2} \frac{x}{3-x} \ dx = \left ( -3 \ln(3-x) - x \right )_{1}^{2}$

$= -3 \ln(3-2) - 2 - \left ( -3 \ln(3-1) - 1 \right )$

= 1.08

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #108 : Definite Integrals

Example Question #109 : Definite Integrals

Example Question #110 : Definite Integrals

Example Question #111 : Definite Integrals

Example Question #112 : Definite Integrals

Example Question #113 : Definite Integrals

Example Question #114 : Definite Integrals

Example Question #115 : Definite Integrals

Example Question #116 : Definite Integrals

Example Question #117 : Definite Integrals

All Calculus 2 Resources

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