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Calculus 1 : Midpoint Riemann Sums

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Example Questions

Example Question #121 : How To Find Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate

$\int_0^{0.15} \tan(\cos(\sin(x^3)))$ using three midpoints.

Possible Answers:

0.102

2.119

0.234

0.008

3.457

Correct answer:

0.234

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 evaluating $\int_0^{0.15} \tan(\cos(\sin(x^3)))$

So the interval is [0,0.15] , the subintervals have length $\frac{0.15-0}{3}=0.05$ , and since we are using the midpoints of each interval, the x-values are [0.025,0.075,0.125]

${\int_0^{0.15} \tan(\cos(\sin(x^3))) \approx (0.05)[\tan(\cos(\sin(0.025^3)))+\tan(\cos(\sin(0.075^3)))+\tan(\cos(\sin(0.125^3)))}]$

${\int_0^{0.15} \tan(\cos(\sin(x^3))) \approx 0.234$

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Example Question #121 : Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate $\int_{10}^{40}e^{\tan(x)}dx$ using three midpoints.

Possible Answers:

17.55

49.16

73.54

28.09

1.08

Correct answer:

49.16

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 $\int_{10}^{40}e^{\tan(x)}dx$

So the interval is [10,40] , the subintervals have length $\frac{40-10}{3}=10$ , and since we are using the midpoints of each interval, the x-values are [15,25,35]

$\int_{10}^{40}e^{\tan(x)}dx\approx (10)[e^{\tan(15)}+e^{\tan(25)}+e^{\tan(35)}]$

$\int_{10}^{40}e^{\tan(x)}dx\approx 49.16$

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Example Question #123 : How To Find Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate $\int_{-4}^{4} \tan(x^3+x+1)dx$ using two midpoints.

Possible Answers:

-99.99

1.05

-40.67

-901.99

11.63

Correct answer:

-901.99

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 $\int_{-4}^{4} \tan(x^3+x+1)dx$

So the interval is [-4,4] , the subintervals have length $\frac{4-(-4)}{2}=4$ , and since we are using the midpoints of each interval, the x-values are [-2,2]

$\int_{-4}^{4} \tan(x^3+x+1)dx \approx (4)[\tan((-2)^3+(-2)+1)+\tan(2^3+2+1)]$

$\int_{-4}^{4} \tan(x^3+x+1)dx \approx -901.99$

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Example Question #124 : How To Find Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate using three midpoints the integral $\int_{-1}^{5} \frac{1}{\cos^2(x^2)}dx$

Possible Answers:

6.14

4.43

12.28

8.86

9.36

Correct answer:

8.86

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 $\int_{-1}^{5} \frac{1}{\cos^2(x^2)}dx$

So the interval is [-1,5] , the subintervals have length $\frac{5-(-1)}{3}=2$ , and since we are using the midpoints of each interval, the x-values are [0,2,4]

$\int_{-1}^{5} \frac{1}{\cos^2(x^2)}dx \approx (2)[\frac{1}{\cos^2(0^2)}+\frac{1}{\cos^2(2^2)}+\frac{1}{\cos^2(4^2)}]$

$\int_{-1}^{5} \frac{1}{\cos^2(x^2)}dx \approx 8.86$

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Example Question #125 : How To Find Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate the integral $\int_{0.19\pi}^{0.37\pi} \tan(\tan(x))dx$ using three midpoints.

Possible Answers:

0.013

-0.013

-0.078

-0.041

0.108

Correct answer:

-0.041

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 $\int_{0.19\pi}^{0.37\pi} \tan(\tan(x))dx$

So the interval is $[0.19\pi,0.37\pi]$ , the subintervals have length $\frac{0.37\pi-0.19\pi}{3}=0.06\pi$ , and since we are using the midpoints of each interval, the x-values are $[0.22\pi,0.28\pi,0.34\pi]$

$\int_{0.19\pi}^{0.37\pi} \tan(\tan(x))dx \approx (0.06\pi)[\tan(\tan(0.22\pi))+\tan(\tan(0.28\pi))+\tan(\tan(0.34\pi))]$

$\int_{0.19\pi}^{0.37\pi} \tan(\tan(x))dx \approx -0.041$

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Example Question #122 : Midpoint Riemann Sums

Use midpoint Riemann sums to give an estimate of the integral $\int_{e}^{e^2}\ln(x^2)dx$ using three midpoints.

Possible Answers:

14.82

19.33

11.05

15.64

17.71

Correct answer:

14.82

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 $\int_{e}^{e^2}\ln(x^2)dx$

So the interval is [e,e^2] , the subintervals have length $\frac{e^2-e}{3}$ , and since we are using the midpoints of each interval, the x-values are $[e+\frac{e^2-e}{6},e+\frac{e^2-e}{6}+\frac{e^2-e}{3},e+\frac{e^2-e}{6}+\frac{2(e^2-e)}{3}]$

or, more simply,

$[\frac{e^2+5e}{6},\frac{e^2+e}{2},\frac{5e^2+e}{6}]$

It is possible, of course, to simply use numerical approximations of each of these terms.

$\int_{e}^{e^2}\ln(x^2)dx \approx (\frac{e^2-e}{3})[\ln((\frac{e^2+5e}{6})^2)+\ln((\frac{e^2+e}{2})^2)+\ln((\frac{5e^2+e}{6})^2)]$

$\int_{e}^{e^2}\ln(x^2)dx \approx 14.82$

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Example Question #127 : How To Find Midpoint Riemann Sums

Using 4 intervals, calculate the midpoint Riemann sum approximation of the area under the curve defined by $\small f(x) = 3x^2 + 5x + 6$ on the interval $\small [0, 8]$ .

Possible Answers:

$\small 256$

$\small 712$

$\small 380$

$\small 968$

$\small 504$

Correct answer:

$\small 712$

Explanation:

To make 4 intervals, we need 5 boundary lines. In this case, they are the vertical lines at $\small x = 0, 2, 4, 6, 8$ . So the midpoints of each interval are at $\small x = 1, 3, 5, 7$ , respectively.

The area of each region is found by $\small f(x)\cdot \Delta x$ , where $\small \Delta x$ is the width of the interval (here, 2). So we simply need to find the functional value at each $\small x$ -coordinate of the midpoints above, multiply by 2, and sum. Equivalently, we could sum the functional values, then double the sum.

$\small f(1) = 14$

$\small f(3) = 48$

$\small f(5) = 106$

$\small f(7) = 188$

Their sum is $\small 356$ , and twice that is $\small 712$ .

The wrong answer $\small 504$ comes from using $\small x = 0, 2, 4, 6$ as the midpoints.

The wrong answer $\small 968$ comes from using $\small x = 2, 4, 6, 8$ as the midpoints.

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Example Question #122 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{0.2}^{0.5}e^{(\frac{1}{2})^{-\frac{1}{x}}}dx$ using three midpoints.

Possible Answers:

$8.89 \cdot 10^5$

$8.89 \cdot 10^2$

$8.89 \cdot 10^4$

$8.89 \cdot 10$

$8.89 \cdot 10^3$

Correct answer:

$8.89 \cdot 10^5$

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 $\int_{0.2}^{0.5}e^{(\frac{1}{2})^{-\frac{1}{x}}}dx$

So the interval is [0.2,0.5] , the subintervals have length $\frac{0.5-0.2}{3}=0.1$ , and since we are using the midpoints of each interval, the x-values are [0.25,0.35,0.45]

$\int_{0.2}^{0.5}e^{(\frac{1}{2})^{-\frac{1}{x}}}dx \approx (0.1)[e^{(\frac{1}{2})^{-\frac{1}{0.25}}}+e^{(\frac{1}{2})^{-\frac{1}{0.35}}}+e^{(\frac{1}{2})^{-\frac{1}{0.45}}}]$

$\int_{0.2}^{0.5}e^{(\frac{1}{2})^{-\frac{1}{x}}}dx \approx 8.89 \cdot 10^5$

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Example Question #129 : How To Find Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{0.8}^{3.2} \frac{0.5}{tan(x^3)}dx$ using three midpoints.

Possible Answers:

-12.92

-13.39

-8.42

-9.17

-10.33

Correct answer:

-10.33

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 $\int_{0.8}^{3.2} \frac{0.5}{tan(x^3)}dx$

So the interval is [0.8,3.2] , the subintervals have length $\frac{3.2-0.8}{3}=0.8$ , and since we are using the midpoints of each interval, the x-values are [1.2,2.0,2.8]

$\int_{0.8}^{3.2} \frac{0.5}{tan(x^3)}dx\approx (0.8)[\frac{0.5}{tan(1.2^3)}+\frac{0.5}{tan(2^3)}+\frac{0.5}{tan(2.8^3)}]$

$\int_{0.8}^{3.2} \frac{0.5}{tan(x^3)}dx\approx -10.33$

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Example Question #130 : How To Find Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{0.7}^{1.3}\frac{sin(x)}{sin(x^3)}dx$ using three midpoints.

Possible Answers:

0.681

1.903

3.408

0.381

0.442

Correct answer:

0.681

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 $\int_{0.7}^{1.3}\frac{sin(x)}{sin(x^3)}dx$

So the interval is [0.7,1.3] , the subintervals have length $\frac{1.3-0.7}{3}=0.2$ , and since we are using the midpoints of each interval, the x-values are [0.8,1.0,1.2]

$\int_{0.7}^{1.3}\frac{sin(x)}{sin(x^3)}dx\approx (0.2)[\frac{sin(0.8)}{sin(0.8^3)}+\frac{sin(1.0)}{sin(1.0^3)}+\frac{sin(1.2)}{sin(1.2^3)}]$

$\int_{0.7}^{1.3}\frac{sin(x)}{sin(x^3)}dx\approx 0.681$

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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 #121 : How To Find Midpoint Riemann Sums

Example Question #121 : Midpoint Riemann Sums

Example Question #123 : How To Find Midpoint Riemann Sums

Example Question #124 : How To Find Midpoint Riemann Sums

Example Question #125 : How To Find Midpoint Riemann Sums

Example Question #122 : Midpoint Riemann Sums

Example Question #127 : How To Find Midpoint Riemann Sums

Example Question #122 : Midpoint Riemann Sums

Example Question #129 : How To Find Midpoint Riemann Sums

Example Question #130 : How To Find Midpoint Riemann Sums

All Calculus 1 Resources

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