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Calculus 2 : Calculus II

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

Example Questions

Example Question #138 : Definite Integrals

$\int \frac{5x^2-x+10}{5}dx$

Possible Answers:

$\frac{x^3}{3}-\frac{x^2}{11}+2x+C$

$\frac{x^3}{3}-\frac{x^2}{10}+2x+C$

$\frac{x^3}{3}-\frac{x^2}{10}-2x$

$\frac{x^3}{4}-\frac{x^2}{10}+2x+C$

$\frac{x^3}{3}-\frac{x^2}{10}+2x$

Correct answer:

$\frac{x^3}{3}-\frac{x^2}{10}+2x+C$

Explanation:

First off, chop up the fraction into three separate terms:

$\int x^2-\frac{1}{5}x+2dx$

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

$\frac{x^3}{3}-\frac{1}{5}(\frac{x^2}{2})+2x$

Simplify and add C because it is an indefinite integral:

$\frac{x^3}{3}-\frac{x^2}{10}+2x+C$

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Example Question #139 : Definite Integrals

$\int_{1}^{5}\frac{7+x}{x}dx$

Possible Answers:

15.8661

15.2661

15.3886

15.2689

12.2661

Correct answer:

15.2661

Explanation:

First off, chop up the fraction into two separate terms:

$\int \frac{7}{x}+1dx$

Now integrate:

7lnx+x

Evaluate at 5 and then 1. Subtract the results.

(7ln5+5)-(7ln1+1)

Round to four places:

15.2661

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Example Question #140 : Definite Integrals

Integrate $f(x)=\frac {x^2}{\sqrt{x}}$ from [1,4] .

Possible Answers:

$\frac {-62}{5}$

$\frac {60}{7}$

$\frac {62}{5}$

None of the Above

Correct answer:

$\frac {62}{5}$

Explanation:

Step 1: Rewrite the denominator as x to a certain power. A square root means that the exponent will have a value of $\frac {1}{2}$ .

We get $x^{\frac {1}{2}}$ .

Step 2: Divide the numerator and denominator of the function. When we divide terms with exponents, we must make sure that the bases are the same. Both bases are , so let's continue. When you divide exponents, you are subtracting them. You subtract the bottom exponent FROM the top exponent.
We get $2-\frac {1}{2}$ . We will convert the into a fraction with denominator . By default, the denominator is . We will multiply the numerator and denominator by 2, so the new fraction is $\frac {4}{2}$ .

Step 3: Subtract the converted fraction (exponent of top) and the exponent of the bottom.

$\frac {4}{2}-\frac {1}{2}=\frac {4-1}{2}=\frac {3}{2}$ .

The exponent of the term is $\frac {3}{2}$ .

Step 4: Integrate...
When we Integrate, we add to the exponent. We then divide the new function with a new exponent by that new exponent.

So, we get: $\frac {x^{\frac {3}{2}+1}}{\frac {3}{2}+1}$ .

Evaluate $\frac {3}{2}+1$ and then rewrite..
$\frac {3}{2}+1=\frac {3}{2}+\frac {2}{2}=\frac {3+2}{2}=\frac {5}{2}$

We get: $\frac {x^{\frac {5}{2}}}{\frac {5}{2}}$

Step 5: Evaluate the upper and the lower limit:

If x=4:

$\frac {4^{\frac {5}{2}}}{\frac {5}{2}}=\frac {\sqrt{4^5}}{\frac {5}{2}}=\frac {\sqrt{1024}}{\frac {5}{2}}=\frac {32}{\frac {5}{2}}$
Flip the denominator and multiply:

$32 \cdot \frac {2}{5}=\frac {64}{5}$

If x=1
$\frac{1^\frac{5}{2}}{\frac{5}{2}}=\frac{1}{\frac{5}{2}}=\frac{1}{1}\times\frac{2}{5}=\frac{2}{5}$

Step 6:

Subtract lower limit from the upper limit:

$\frac{64}{5}-\frac{2}{5}=\frac{62}{5}$

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Example Question #141 : Definite Integrals

Evaluate $\int_{-3}^3 \sqrt{9-x^2} dx$ .

Possible Answers:

The integral cannot be evaluated.

$\frac{9\pi}{2}$

$\frac{7\pi}{2}$

$\pi$

Correct answer:

$\frac{9\pi}{2}$

Explanation:

This integral cannot be evaluated using the Fundamental Theorem of Calculus, since an antiderivative does not exist for $\sqrt{9-x^2}$ . Instead, we need to find the area under the curve $y=\sqrt{9-x^2}$ bounded by the axis.

$y=\sqrt{9-x^2}$ from -3<x<3 , is the graph of the upper half of the circle y^2+x^2=3^2 . The area of this upper half is

$\frac{\pi r^2}{2} = \frac{\pi(3)^2}{2} = \frac{9\pi}{2}$ .

Hence $\int_{-3}^3 \sqrt{9-x^2}dx=\frac{9\pi}{2}$ .

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Example Question #141 : Definite Integrals

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

Possible Answers:

3.3274

3.3863

2.3863

1.5863

3.3898

Correct answer:

3.3863

Explanation:

First, chop up the fraction into two separate fractions:

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

Now, integrate:

$2ln\left | x \right |+x$

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

2ln(2)+2-(2ln(0)+0)=3.3863 .

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Example Question #142 : Definite Integrals

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

Possible Answers:

$\frac{74}{3}$

$\frac{94}{3}$

$\frac{26}{3}$

$\frac{22}{3}$

$\frac{14}{3}$

Correct answer:

$\frac{74}{3}$

Explanation:

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

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

Simplify to get:

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

Now, evaluate at 3 and then at 1. Subtract the results:
$(9+18)-(\frac{1}{3}+2)=27-(2\frac{1}{3})=\frac{74}{3}$

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Example Question #143 : Definite Integrals

$\int_{0}^{3}x^{\frac{4}{5}}dx$

Possible Answers:

4.0137

4.0283

3.0137

4.0297

4.0172

Correct answer:

4.0137

Explanation:

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

$\frac{x^{\frac{9}{5}}}{\frac{9}{5}}$

Simplify to get:

$\frac{5}{9}x^{\frac{9}{5}}$

Now, evaluate at 3 and then 0. Subtract the results:

$(\frac{5}{9}(3)^{\frac{9}{5}})-(\frac{5}{9}(0)^{\frac{9}{5}})=4.0137$

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Example Question #501 : Integrals

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

Possible Answers:

Correct answer:

Explanation:

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

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

Now, evaluate at 2 and then 1:

(8+8-2)-(1+2-1)=14-2=12 .

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Example Question #145 : Definite Integrals

$\int_{0}^{1}x^{\frac{1}{7}}dx$

Possible Answers:

$\frac{1}{4}$

$\frac{1}{8}$

$\frac{2}{3}$

$\frac{7}{8}$

$\frac{9}{8}$

Correct answer:

$\frac{7}{8}$

Explanation:

Recall that when integrating, you must raise the exponent by 1 and also put that result on the denominator:

$\frac{x^{\frac{8}{7}}}{\frac{8}{7}}$

Simplify to get:
$\frac{7}{8}x^{\frac{8}{7}}$

Now, evaluate at 1 and then at 0:

$(\frac{7}{8}1^{\frac{8}{7}})-(\frac{7}{8}0^{\frac{8}{7}})=\frac{7}{8}$

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Example Question #146 : Definite Integrals

$\int_{1}^{2}x^3+4x^2+xdx$

Possible Answers:

$\frac{175}{8}$

$\frac{75}{12}$

$\frac{173}{12}$

$\frac{175}{12}$

$\frac{179}{12}$

Correct answer:

$\frac{175}{12}$

Explanation:

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

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

Next, evaluate at 2 and then 1. Subtract the results:
$(4+\frac{32}{3}+2)-(\frac{1}{4}+\frac{4}{3}+\frac{1}{2})$

Simplify:

$(6+\frac{32}{3})-(\frac{50}{24})=\frac{175}{12}$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #138 : Definite Integrals

Example Question #139 : Definite Integrals

Example Question #140 : Definite Integrals

Example Question #141 : Definite Integrals

Example Question #141 : Definite Integrals

Example Question #142 : Definite Integrals

Example Question #143 : Definite Integrals

Example Question #501 : Integrals

Example Question #145 : Definite Integrals

Example Question #146 : Definite Integrals

All Calculus 2 Resources

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