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Calculus 1 : Functions

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Example Questions

Example Question #141 : How To Find Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate the average of $f(x)=x^{x^2}$ over the interval (1,1.8) using four midpoints.

Possible Answers:

2.451

0.490

1.961

0.980

3.922

Correct answer:

2.451

Explanation:

To find the average of a function over a given interval of values [a,b] , the most precise method is to use an integral as follows:

$\frac{1}{b-a}\int_{a}^{b} f(x)dx$

Now for functions that are difficult or impossible to integrate,a Riemann sum can be used to approximate the value. A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length equal to the subinterval length $\frac{b-a}{n}$ , and variable heights f(x_i) , which depend on the function value at a given point x_i .

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

$\frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

$\frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))$

We're asked to approximate the average of $f(x)=x^{x^2}$ over the interval (1,1.8)

The subintervals have length $\frac{1.8-1}{4}=0.2$ , and since we are using the midpoints of each interval, the x-values are [1.1,1.3,1.5,1.7]

$\frac{1}{0.8}\int_{1}^{1.8}x^{x^2}dx\approx (\frac{1}{4})[1.1^{1.1^2}+1.3^{1.3^2}+1.5^{1.5^2}+1.7^{1.7^2}]$

$\frac{1}{0.8}\int_{1}^{1.8}x^{x^2}dx\approx 2.451$

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Example Question #142 : How To Find Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate the average of $f(x)=1.5^{x^2+x}$ over the interval (2,3.2) using three midpoints.

Possible Answers:

63.866

25.546

76.639

98.223

40.802

Correct answer:

63.866

Explanation:

To find the average of a function over a given interval of values [a,b] , the most precise method is to use an integral as follows:

$\frac{1}{b-a}\int_{a}^{b} f(x)dx$

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

$\frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

$\frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))$

We're asked to approximate the average of $f(x)=1.5^{x^2+x}$ over the interval (2,3.2)

The subintervals have length $\frac{3.2-2}{3}=0.4$ , and since we are using the midpoints of each interval, the x-values are [2.2,2.6,3.0]

$\frac{1}{1.2}\int_{2}^{3.2}1.5^{x^2+x}dx\approx (\frac{1}{3})[1.5^{2.2^2+2.2}+1.5^{2.6^2+2.6}+1.5^{3^2+3}]$

$\frac{1}{1.2}\int_{2}^{3.2}1.5^{x^2+x}dx\approx 63.866$

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Example Question #141 : Functions

Utilize the method of midpoint Riemann sums to approximate the average of $f(x)=(\frac{1}{4})^{cos(x)}$ over the interval (3,3.6) using three midpoints.

Possible Answers:

3.591

2.824

2.318

3.863

4.996

Correct answer:

3.863

Explanation:

To find the average of a function over a given interval of values [a,b] , the most precise method is to use an integral as follows:

$\frac{1}{b-a}\int_{a}^{b} f(x)dx$

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

$\frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

$\frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))$

We're asked to approximate the average of $f(x)=(\frac{1}{4})^{cos(x)}$ over the interval (3,3.6)

The subintervals have length $\frac{3.6-3}{3}=0.2$ , and since we are using the midpoints of each interval, the x-values are [3.1,3.3,3.5]

$\frac{1}{3.6-3}\int_{3}^{3.6}(\frac{1}{4})^{cos(x)}dx\approx (\frac{1}{3})[(\frac{1}{4})^{cos(3.1)}+(\frac{1}{4})^{cos(3.3)}+(\frac{1}{4})^{cos(3.5)}]$

$\frac{1}{0.6}\int_{3}^{3.6}(\frac{1}{4})^{cos(x)}dx\approx 3.863$

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Example Question #142 : Functions

Utilize the method of midpoint Riemann sums to approximate the average of $f(x)=2^{csc(x)}$ over the interval (2,5) using three midpoints.

Possible Answers:

0.240

1.975

3.815

1.272

2.208

Correct answer:

1.272

Explanation:

To find the average of a function over a given interval of values [a,b] , the most precise method is to use an integral as follows:

$\frac{1}{b-a}\int_{a}^{b} f(x)dx$

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

$\frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

$\frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))$

We're asked to approximate the average of $f(x)=2^{csc(x)}$ over the interval (2,5)

The subintervals have length $\frac{5-2}{3}=1$ , and since we are using the midpoints of each interval, the x-values are [2.5,3.5,4.5]

$\frac{1}{5-2}\int_{2}^{5}2^{csc(x)}dx\approx (\frac{1}{3})[2^{csc(2.5)}+2^{csc(3.5)}+2^{csc(4.5)}]$

$\frac{1}{3}\int_{2}^{5}2^{csc(x)}dx\approx 1.272$

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Example Question #142 : How To Find Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate the average of $f(x)=e^{tan(x)}$ over the interval (0,1.5) using three midpoints.

Possible Answers:

15.303

7.271

8.036

6.199

12.054

Correct answer:

8.036

Explanation:

To find the average of a function over a given interval of values [a,b] , the most precise method is to use an integral as follows:

$\frac{1}{b-a}\int_{a}^{b} f(x)dx$

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

$\frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

$\frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))$

We're asked to approximate the average of $f(x)=e^{tan(x)}$ over the interval (0,1.5)

The subintervals have length $\frac{1.5-0}{3}=0.5$ , and since we are using the midpoints of each interval, the x-values are [0.25,0.75,1.25]

$\frac{1}{1.5-0}\int_{0}^{1.5}e^{tan(x)}dx\approx (\frac{1}{3})[e^{tan(0.25)}+e^{tan(0.75)}+e^{tan(1.25)}]$

$\frac{1}{1.5}\int_{0}^{1.5}e^{tan(x)}dx\approx 8.036$

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Example Question #143 : How To Find Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate the average of $f(x)=sec^3(\frac{1}{x})$ over the interval (2,2.6) using three midpoints.

Possible Answers:

0.543

1.348

0.809

2.104

0.270

Correct answer:

1.348

Explanation:

To find the average of a function over a given interval of values [a,b] , the most precise method is to use an integral as follows:

$\frac{1}{b-a}\int_{a}^{b} f(x)dx$

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

$\frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

$\frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))$

We're asked to approximate the average of $f(x)=sec^3(\frac{1}{x})$ over the interval (2,2.6)

The subintervals have length $\frac{2.6-2}{3}=0.2$ , and since we are using the midpoints of each interval, the x-values are [2.1,2.3,2.5]

$\frac{1}{2.6-2}\int_{2}^{2.6}sec^3(\frac{1}{x})dx\approx (\frac{1}{3})[sec^3(\frac{1}{2.1})+sec^3(\frac{1}{2.3})+sec^3(\frac{1}{2.5})]$

$\frac{1}{0.6}\int_{2}^{2.6}sec^3(\frac{1}{x})dx\approx 1.348$

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Example Question #141 : Functions

Utilize the method of midpoint Riemann sums to approximate the average of f(x)=csc^2(x^3) over the interval (1,7) using three midpoints.

Possible Answers:

6.928

3.087

8.534

1.422

2.845

Correct answer:

1.422

Explanation:

To find the average of a function over a given interval of values [a,b] , the most precise method is to use an integral as follows:

$\frac{1}{b-a}\int_{a}^{b} f(x)dx$

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

$\frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

$\frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))$

We're asked to approximatethe average of f(x)=csc^2(x^3) over the interval (1,7)

The subintervals have length $\frac{7-1}{3}=2$ , and since we are using the midpoints of each interval, the x-values are [2,4,6]

$\frac{1}{7-1}\int_{1}^{7}csc^2(x^3)dx\approx (\frac{1}{3})[csc^2(2^3)+csc^2(4^3)+csc^2(6^3)]$

$\frac{1}{6}\int_{1}^{7}csc^2(x^3)dx\approx 1.422$

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Example Question #144 : How To Find Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate the average of $f(x)=\frac{sin(x^2)}{tan(x^3)}$ over the interval (0.3,0.306) using three midpoints.

Possible Answers:

0.007

1.623

0.020

3.295

2.809

Correct answer:

3.295

Explanation:

To find the average of a function over a given interval of values [a,b] , the most precise method is to use an integral as follows:

$\frac{1}{b-a}\int_{a}^{b} f(x)dx$

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

$\frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

$\frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))$

We're asked to approximate the average of $f(x)=\frac{sin(x^2)}{tan(x^3)}$ over the interval (0.3,0.306)

The subintervals have length $\frac{0.306-0.3}{3}=0.002$ , and since we are using the midpoints of each interval, the x-values are [3.001,3.003,3.005]

$\frac{1}{0.306-0.3}\int_{0.3}^{0.306}\frac{sin(x^2)}{tan(x^3)}dx\approx (\frac{1}{3})[\frac{sin(0.301^2)}{tan(0.301^3)}+\frac{sin(0.303^2)}{tan(0.303^3)}+\frac{sin(0.305^2)}{tan(0.305^3)}]$

$\frac{1}{0.006}\int_{0.3}^{0.306}\frac{sin(x^2)}{tan(x^3)}dx\approx 3.295$

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Example Question #149 : How To Find Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate the average of $f(x)=5^{\sqrt{x}}$ over the interval (0.2,0.8) using three midpoints.

Possible Answers:

1.876

0.625

4.067

3.126

9.379

Correct answer:

3.126

Explanation:

To find the average of a function over a given interval of values [a,b] , the most precise method is to use an integral as follows:

$\frac{1}{b-a}\int_{a}^{b} f(x)dx$

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

$\frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

$\frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))$

We're asked to approximate the average of $f(x)=5^{\sqrt{x}}$ over the interval (0.2,0.8)

The subintervals have length $\frac{0.8-0.2}{3}=0.2$ , and since we are using the midpoints of each interval, the x-values are [0.3,0.5,0.7]

$\frac{1}{0.8-0.2}\int_{0.2}^{0.8}5^{\sqrt{x}}dx\approx (\frac{1}{3})[5^{\sqrt{0.3}}+5^{\sqrt{0.5}}+5^{\sqrt{0.7}}]$

$\frac{1}{0.6}\int_{0.2}^{0.8}5^{\sqrt{x}}dx\approx 3.126$

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Example Question #145 : How To Find Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate the average of f(x)=tan^4(x) over the interval (1,4) using three midpoints.

Possible Answers:

13180.56

17235.24

21022.55

39541.67

9088.32

Correct answer:

13180.56

Explanation:

To find the average of a function over a given interval of values [a,b] , the most precise method is to use an integral as follows:

$\frac{1}{b-a}\int_{a}^{b} f(x)dx$

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

$\frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

$\frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))$

We're asked to approximate the average of f(x)=tan^4(x) over the interval (1,4)

The subintervals have length $\frac{4-1}{3}=1$ , and since we are using the midpoints of each interval, the x-values are [1.5,2.5,3.5]

$\frac{1}{4-1}\int_{1}^{4}tan^4(x)dx\approx (\frac{1}{3})[tan^4(1.5)+tan^4(2.5)+tan^4(3.5)]$

$\frac{1}{3}\int_{1}^{4}tan^4(x)dx\approx 13180.56$

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #141 : How To Find Midpoint Riemann Sums

Example Question #142 : How To Find Midpoint Riemann Sums

Example Question #141 : Functions

Example Question #142 : Functions

Example Question #142 : How To Find Midpoint Riemann Sums

Example Question #143 : How To Find Midpoint Riemann Sums

Example Question #141 : Functions

Example Question #144 : How To Find Midpoint Riemann Sums

Example Question #149 : How To Find Midpoint Riemann Sums

Example Question #145 : How To Find Midpoint Riemann Sums

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