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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 #531 : Integrals

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

Possible Answers:

Correct answer:

Explanation:

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

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

Now, evaluate at 3 and then 1. Subtract the results:

$(27-\frac{9}{2}+3)-(1-\frac{1}{2}+1)=28-4=24$

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Example Question #182 : Finding Integrals

$\int_{0}^{2}x^{\frac{3}{2}}dx$

Possible Answers:

3.2627

2.2692

2.4727

2.2197

2.2627

Correct answer:

2.2627

Explanation:

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{x^{\frac{5}{2}}}{\frac{5}{2}}$

Simplify:

$\frac{2}{5}x^{\frac{5}{2}}$

Now evaluate at 2 and then 0. Subtract the results:

$\frac{2}{5}(2^{\frac{5}{2}})-\frac{2}{5}(o^{\frac{5}{2}})=2.2627-0=2.2627$

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Example Question #168 : Definite Integrals

$\int_{1}^{3}3x^2+2x+2dx$

Possible Answers:

Correct answer:

Explanation:

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

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

Simplify:

x^3+x^2+2x

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

(27+9+6)-(1+1+2)=42-4=38

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Example Question #184 : Finding Integrals

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

Possible Answers:

$\frac{51}{8}$

$\frac{51}{4}$

$\frac{39}{2}$

$\frac{51}{2}$

$\frac{53}{2}$

Correct answer:

$\frac{51}{2}$

Explanation:

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

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

Now, evaluate at 4 and then 1. Subtract the results:

$(\frac{64}{3}-8+16)-(\frac{1}{3}-\frac{1}{2}+4)$

Simplify to get your answer:

$4+\frac{129}{6}=\frac{51}{2}$

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Example Question #531 : Integrals

$\int_{0}^{2}x^4+x-1dx$

Possible Answers:

$\frac{32}{3}$

$\frac{36}{5}$

$\frac{32}{9}$

$\frac{32}{5}$

$\frac{31}{5}$

Correct answer:

$\frac{32}{5}$

Explanation:

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{x^5}{5}+\frac{x^2}{2}-x$

Evaluate at 2 and then 0. Subtract the results:

$(\frac{32}{5}+2-2)-0=\frac{32}{5}$

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Example Question #186 : Finding Integrals

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

Possible Answers:

$\frac{19}{3}$

$\frac{20}{3}$

$\frac{17}{3}$

$\frac{16}{3}$

$\frac{19}{4}$

Correct answer:

$\frac{19}{3}$

Explanation:

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{4x^3}{3}-3x$

Now, evaluate at 2 and then at 1. Subtract the results:

$(\frac{32}{3}-6)-(\frac{4}{3}-3)$

Simplify to get your answer:

$\frac{28}{3}-3=\frac{19}{3}$

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Example Question #174 : Definite Integrals

$\int_{1}^{3} 3x^4+7x+3$

Possible Answers:

180

177

$-\frac {896}{5}$

$\frac {896}{5}$

Correct answer:

$\frac {896}{5}$

Explanation:

Step 1: Integrate:

$\int_{1}^{3}3x^4+7x+3\rightarrow \frac {3x^5}{5}+\frac {7x^2}{2}+3x|_{1}^{3}$

Step 2: Evaluate at the upper limit:

Plug in x=3 .

$\frac {3(3)^5}{5}+\frac {7(3)^2}{2}+3(3)=\frac {3^6}{5}+\frac {63}{2}+9=186.3$

Step 3: Evaluate at the lower limit:

Plug in x=1 .

$\frac {3(1)^5}{5}+\frac {7(1)^2}{2}+3(1)=\frac {3}{5}+\frac {7}{2}+3=7.1$

Step 4: Take the valuation at the lower limit and subtract it FROM the upper limit:

$186.3-7.1=179.2=\frac {896}{5}$

The integration of 3x^4+7x+3 is $\frac {896}{5}$

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Example Question #175 : Definite Integrals

Evaluate:

$\int_{0}^{2} 2x^3+3x^2+3x$

Possible Answers:

Correct answer:

Explanation:

Step 1: Take the antiderivative of each term:

$2x^3 \rightarrow 2 \cdot \frac {x^4}{4}=\frac {x^4}{2}$

$3x^2=3 \cdot \frac {x^3}{3}=x^3$

$3x\rightarrow 3 \cdot \frac {x^2}{2}=\frac {3x^2}{2}$

Step 2: Put all the antiderivatives in step 1 together based on the signs in the integral...

$\frac {x^4}{2}+x^3+\frac {3x^2}{2}$

Step 3: Plug in the upper and lower limits:

Upper Limit is , lower limit is .

Plug in x=2 :

$\frac {2^4}{2}+2^3+\frac {3(2)^2}{2}=2^3+2^3+3(2)=8+8+6=22$

Plug in x=0 . Since all terms have , the value will be

Step 4: Subtract the value of the lower limit from the upper limit:

22-0=22

The value of this integral is .

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Example Question #187 : Finding Integrals

$\int_{0}^{2}4x^2-xdx$

Possible Answers:

$\frac{22}{3}$

$\frac{31}{3}$

$\frac{26}{3}$

$\frac{25}{3}$

$\frac{29}{3}$

Correct answer:

$\frac{26}{3}$

Explanation:

First, integrate this expression. Remember to raise the exponent by 1 and then also put that result on the denominator:

$\frac{4x^3}{3}-\frac{x^2}{2}$

Now, evaluate at 2 and then 0. Subtract the results:

$(\frac{32}{3}-2)-(0)$

Simplify to get your answer:

$\frac{32}{3}-\frac{6}{3}=\frac{26}{3}$

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Example Question #172 : Definite Integrals

$\int_{0}^{2}4\sqrt{x}dx$

Possible Answers:

7.5499

7.5425

7.5431

6.5425

7.5295

Correct answer:

7.5425

Explanation:

First, take the 4 outside of the integral sign and rewrite the radical as a fractional exponent. It's easier to visualize that when integrating:

$4\int x^{\frac{1}{2}}dx$

Now integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

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

Simplify and multiply by the 4 that you took out:

$4(\frac{2}{3})x^{\frac{3}{2}}=\frac{8}{3}x^{\frac{3}{2}}$

Now evaluate at 2 and then 0. Subtract the results:

$\frac{8}{3}2^{\frac{3}{2}}-\frac{8}{3}0^{\frac{3}{2}}=7.5425$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #531 : Integrals

Example Question #182 : Finding Integrals

Example Question #168 : Definite Integrals

Example Question #184 : Finding Integrals

Example Question #531 : Integrals

Example Question #186 : Finding Integrals

Example Question #174 : Definite Integrals

Example Question #175 : Definite Integrals

Example Question #187 : Finding Integrals

Example Question #172 : Definite Integrals

All Calculus 2 Resources

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