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Calculus 2 : Definite Integrals

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Example Questions

Example Question #131 : Definite Integrals

Evaluate.

$\int_{6}^{7} \frac{x-4}{x-5} \ dx$

Possible Answers:

45.2

Answer not listed.

18.3

16.7

3.5

Correct answer:

Answer not listed.

Explanation:

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

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

The antiderivative is $F(x) =\ln(x-5) + x$ .

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

$\int_{6}^{7} \frac{x-4}{x-5} \ dx = \left (\ln(x-5) + x \right )_{6}^{7}$

$= \ln(7-5) + 7 - \left ( \ln(6-5) + 6 \right )$

= 13.7

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Example Question #132 : Definite Integrals

Evaluate.

$\int_{0}^{1} \cos(2x-9) \ dx$

Possible Answers:

-4.2

1.54

3.8

0.02

Answer not listed.

Correct answer:

0.02

Explanation:

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

In this case, $f(x) = \cos(2x-9)$ .

The antiderivative is $F(x) = \frac{\sin(2x-9)}{2}$ .

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

$\int_{0}^{1} \cos(2x-9) \ dx = \left ( \frac{\sin(2x-9)}{2} \right )_{0}^{1}$

$= \frac{\sin(2(1)-9)}{2} - \left ( \frac{\sin(2(0)-9)}{2} \right )$

= 0.02

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Example Question #133 : Definite Integrals

Evaluate.

$\int_{1}^{2} -\frac{\sin(2x)}{2} \ dx$

Possible Answers:

2.04

Answer not listed.

1.38

-2.03

-0.06

Correct answer:

-0.06

Explanation:

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

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

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

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

$\int_{1}^{2} -\frac{\sin(2x)}{2} \ dx= \left ( \frac{\cos(2x)}{4} \right )_{1}^{2}$

$= \frac{\cos(2(2))}{4} - \left ( \frac{\cos(2(1))}{4} \right )$

= -0.06

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Example Question #134 : Definite Integrals

Evaluate.

$\int_{2}^{3}\sec^2 (12x-1)\ dx$

Possible Answers:

Answer not listed.

2.74

4.26

0.94

0.02

Correct answer:

0.02

Explanation:

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

In this case, $f(x) =\sec^2 (12x-1)$ .

The antiderivative is $F(x) = \frac{\tan(12x-1)}{12}$ .

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

$\int_{2}^{3}\sec^2 (12x-1)\ dx = \left ( \frac{\tan(12x-1)}{12} \right )_{2}^{3}$

$= \frac{\tan(12(3)-1)}{12} - \left ( \frac{\tan(12(2)-1)}{12} \right )$

= 0.02

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Example Question #135 : Definite Integrals

$\int_{0}^{2}x^3+4x^2-x+1dx$

Possible Answers:

$\frac{41}{3}$

$\frac{38}{3}$

$\frac{50}{3}$

$\frac{43}{3}$

$\frac{44}{3}$

Correct answer:

$\frac{44}{3}$

Explanation:

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

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

Now, evaluate at 2 and then 0. Subtract the results:
$(\frac{16}{4}+\frac{32}{3}-\frac{4}{2}+2)-0=4+\frac{32}{3}-2+2=\frac{44}{3}$

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Example Question #136 : Definite Integrals

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

Possible Answers:

3.5069

1.8058

2.3456

1.2950

1.3506

Correct answer:

1.3506

Explanation:

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

$\frac{x^{\frac{7}{4}}}{\frac{7}{4}}$

Simplify:

$\frac{4}{7}(x^{\frac{7}{4}})$

Evaluate at 2 and then 1. Subtract the results:

$(\frac{4}{7}(2^{\frac{7}{4}}))-(\frac{4}{7}(1^{\frac{7}{4}}))$

Round to four places:

1.9220-.5714=1.3506

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Example Question #137 : Definite Integrals

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

Possible Answers:

9.4315

14.4328

12.4315

14.4315

14.4372

Correct answer:

14.4315

Explanation:

First, chop up the fraction into two separate terms:

$\int \frac{5}{x}+xdx$

Now, integrate:

$5ln\left | x \right |+\frac{x^2}{2}$

Evaluate at 4 and then at 1. Subtract the results:

$(5ln4+8)-(5ln1+\frac{1}{2})$

Round to four places:
14.9315-.5=14.4315

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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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Calculus 2 : Definite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #131 : Definite Integrals

Example Question #132 : Definite Integrals

Example Question #133 : Definite Integrals

Example Question #134 : Definite Integrals

Example Question #135 : Definite Integrals

Example Question #136 : Definite Integrals

Example Question #137 : Definite Integrals

Example Question #138 : Definite Integrals

Example Question #139 : Definite Integrals

Example Question #140 : Definite Integrals

All Calculus 2 Resources

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