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Calculus 1 : Midpoint Riemann Sums

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Example Questions

Example Question #161 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{0}^{3}x^3sin(x^3)dx$ using four midpoints.

Possible Answers:

-8.224

-6.802

-7.345

-9.070

-5.333

Correct answer:

-6.802

Explanation:

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 $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{0}^{3}x^3sin(x^3)dx$

So the interval is [0,3] , the subintervals have length $\frac{3-0}{4}=0.75$ , and since we are using the midpoints of each interval, the x-values are [0.375,1.125,1.875,2.625]

${\int_{0}^{3}x^3sin(x^3)dx\approx (0.75)[0.375^3sin(0.375^3)+1.125^3sin(1.125^3)+1.875^3sin(1.875^3)+2.625^3sin(2.625^3)]}$

$\int_{0}^{3}x^3sin(x^3)dx\approx -6.802$

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Example Question #162 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{0.1}^{0.22}\frac{tan(x)}{x^2}dx$ using three midpoints.

Possible Answers:

0.790

9.009

1.998

13.212

19.745

Correct answer:

0.790

Explanation:

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 $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{0.1}^{0.22}\frac{tan(x)}{x^2}dx$

So the interval is [0.1,0.22] , the subintervals have length $\frac{0.22-0.1}{3}=0.04$ , and since we are using the midpoints of each interval, the x-values are [0.12,0.16,0.2]

$\int_{0.1}^{0.22}\frac{tan(x)}{x^2}dx\approx (0.04)[\frac{tan(0.12)}{0.12^2}+\frac{tan(0.16)}{0.16^2}+\frac{tan(0.2)}{0.2^2}]$

$\int_{0.1}^{0.22}\frac{tan(x)}{x^2}dx\approx 0.790$

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Example Question #163 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{1.5}^{1.8}\frac{x^4}{tan(x)}dx$ using three midpoints.

Possible Answers:

-3.111

-0.513

-0.044

-1.097

-0.217

Correct answer:

-0.217

Explanation:

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 $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{1.5}^{1.8}\frac{x^4}{tan(x)}dx$

So the interval is [1.5,1.8] , the subintervals have length $\frac{1.8-1.5}{3}=0.1$ , and since we are using the midpoints of each interval, the x-values are [1.55,1.65,1.75]

$\int_{1.5}^{1.8}\frac{x^4}{tan(x)}dx\approx (0.1)[\frac{1.55^4}{tan(1.55)}+\frac{1.65^4}{tan(1.65)}+\frac{1.75^4}{tan(1.75)}]$

$\int_{1.5}^{1.8}\frac{x^4}{tan(x)}dx\approx -0.217$

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Example Question #164 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{7.8}^{7.86}\frac{x^3}{2tan(x)}dx$ using three midpoints.

Possible Answers:

17.203

0.344

-0.503

-12.089

1.769

Correct answer:

0.344

Explanation:

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 $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{7.8}^{7.86}\frac{x^3}{2tan(x)}dx$

So the interval is [7.8,7.86] , the subintervals have length $\frac{7.86-7.8}{3}=0.02$ , and since we are using the midpoints of each interval, the x-values are [7.81,7.83,7.85]

$\int_{7.8}^{7.86}\frac{x^3}{2tan(x)}dx\approx (0.02)[\frac{7.81^3}{2tan(7.81)}+\frac{7.83^3}{2tan(7.83)}+\frac{7.85^3}{2tan(7.85)}]$

$\int_{7.8}^{7.86}\frac{x^3}{2tan(x)}dx\approx 0.344$

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Example Question #165 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{1.5}^{2.1}\frac{x^x}{tan(x)}dx$ using three midpoints.

Possible Answers:

4.462

3.124

-0.513

0.809

-2.565

Correct answer:

-0.513

Explanation:

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 $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{1.5}^{2.1}\frac{x^x}{tan(x)}dx$

So the interval is [1.5,2.1] , the subintervals have length $\frac{2.1-1.5}{3}=0.2$ , and since we are using the midpoints of each interval, the x-values are [1.6,1.8,2]

$\int_{1.5}^{2.1}\frac{x^{x}}{tan(x)}dx\approx (0.2)[\frac{1.6^{1.6}}{tan(1.6)}+\frac{1.8^{1.8}}{tan(1.8)}+\frac{2^{2}}{tan(2)}]$

$\int_{1.5}^{2.1}\frac{x^{x}}{tan(x)}dx\approx -0.513$

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Example Question #166 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{0}^{6}(x!)^{-\sqrt{x}}dx$ using three midpoints.

Possible Answers:

10.303

19.881

2.090

245.277

0.438

Correct answer:

2.090

Explanation:

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 $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{0}^{6}(x!)^{-\sqrt{x}}dx$

So the interval is [0,6] , the subintervals have length $\frac{6-0}{3}=2$ , and since we are using the midpoints of each interval, the x-values are [1,3,5]

$\int_{0}^{6}(x!)^{-\sqrt{x}}dx\approx (2)[(1!)^{-\sqrt{1}}+(3!)^{-\sqrt{3}}+(5!)^{-\sqrt{5}}]$

$\int_{0}^{6}(x!)^{-\sqrt{x}}dx\approx 2.090$

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Example Question #161 : How To Find Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate the average of f(x)=14sin^4(x) over the interval (0,6) using three midpoints.

Possible Answers:

6.287

23.856

37.725

13.373

29.001

Correct answer:

6.287

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)=14sin^4(x) over the interval (0,6)

The subintervals have length $\frac{6-0}{3}=2$ , and since we are using the midpoints of each interval, the x-values are [1,3,5]

$\frac{1}{6}\int_{0}^{6}14sin^4(x)dx\approx (\frac{1}{3})[14sin^4(1)+14sin^4(3)+14sin^4(5)]$

$\frac{1}{6}\int_{0}^{6}14sin^4(x)dx\approx 6.287$

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Example Question #168 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate the average of $f(x)=2^\frac{x}{3}$ over the interval (0,10) using five midpoints.

Possible Answers:

3.895

15.563

27.258

1.556

38.949

Correct answer:

3.895

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 $f(x)=2^\frac{x}{3}$ over the interval (0,10)

The subintervals have length $\frac{10-0}{5}=2$ , and since we are using the midpoints of each interval, the x-values are [1,3,5,7,9]

$\frac{1}{10}\int_{0}^{10}2^{\frac{x}{3}}dx\approx (\frac{1}{5})[2^{\frac{1}{3}}+2^{\frac{3}{3}}+2^{\frac{5}{3}}+2^{\frac{7}{3}}+2^{\frac{9}{3}}]$

$\frac{1}{10}\int_{0}^{10}2^\frac{x}{3}dx\approx 3.895$

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Example Question #161 : How To Find Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate the average of $f(x)=tan(\frac{1}{tan(x)})$ over the interval (0.5,3.5) using three midpoints.

Possible Answers:

1.009

-0.755

-0.214

3.027

-0.643

Correct answer:

-0.214

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 $f(x)=tan(\frac{1}{tan(x)})$ over the interval (0.5,3.5)

The subintervals have length $\frac{3.5-0.5}{3}=1$ , and since we are using the midpoints of each interval, the x-values are [1,2,3]

$\frac{1}{3}\int_{0.5}^{3.5}tan(\frac{1}{tan(x)})dx\approx (\frac{1}{3})[tan(\frac{1}{tan(1)})+tan(\frac{1}{tan(2)})+tan(\frac{1}{tan(3)})]$

$\frac{1}{3}\int_{0.5}^{3.5}tan(\frac{1}{tan(x)})dx\approx -0.214$

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Example Question #1191 : Calculus

Utilize the method of midpoint Riemann sums to approximate the average of f(x)=sec(cos(x^2)) over the interval (1.5,4.5) using three midpoints.

Possible Answers:

1.734

1.543

2.281

4.629

0.578

Correct answer:

1.543

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(cos(x^2)) over the interval (1.5,4.5)

The subintervals have length $\frac{4.5-1.5}{3}=1$ , and since we are using the midpoints of each interval, the x-values are [2,3,4]

$\frac{1}{3}\int_{1.5}^{4.5}sec(cos(x^2))dx\approx (\frac{1}{3})[sec(cos(2^2))+sec(cos(3^2))+sec(cos(4^2))]$

$\frac{1}{3}\int_{1.5}^{4.5}sec(cos(x^2))dx\approx 1.543$

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Calculus 1 : Midpoint Riemann Sums

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #161 : Midpoint Riemann Sums

Example Question #162 : Midpoint Riemann Sums

Example Question #163 : Midpoint Riemann Sums

Example Question #164 : Midpoint Riemann Sums

Example Question #165 : Midpoint Riemann Sums

Example Question #166 : Midpoint Riemann Sums

Example Question #161 : How To Find Midpoint Riemann Sums

Example Question #168 : Midpoint Riemann Sums

Example Question #161 : How To Find Midpoint Riemann Sums

Example Question #1191 : Calculus

All Calculus 1 Resources

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