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Calculus 1 : How to find midpoint Riemann sums

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Example Questions

Example Question #1071 : Calculus

Using the method of midpoint Reimann sums, approximate the integral of the function $f(x)=e^{\frac{1}{x}}$ over the interval x=[1,6] using four midpoints.

Possible Answers:

8.6178

7.1815

10.9372

5.7452

5.1283

Correct answer:

7.1815

Explanation:

The Reimann sum approximation of an integral of a function with subintervals over an interval [a,b] takes the form:

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

Where $\frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are four midpoints, the subintervals have length $\frac{6-1}{4}=1.25$ , and the midpoints are x=[1.625,2.875,4.125,5.375] .

The integral is thus:

$\int_1^6 e^{\frac{1}{x}}dx \approx 1.25[e^{\frac{1}{1.625}}+e^{\frac{1}{2.875}}+e^{\frac{1}{4.125}}+e^{\frac{1}{5.375}}]$

$\int_1^6 e^{\frac{1}{x}}dx \approx 7.1815$

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Example Question #1081 : Calculus

Using the method of midpoint Reimann sums, approximate the area between the curves $f(x)=e^{\frac{1}{x}}$ and $g(x)=e^{x^2}$ over the interval x=[0.5,1] using four midpoints.

Possible Answers:

0.6214

2.4571

0.8952

1.1314

1.5927

Correct answer:

1.1314

Explanation:

The Reimann sum approximation of an integral of a function with subintervals over an interval [a,b] takes the form:

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

Where $\frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are four midpoints, the subintervals have length $\frac{1-0.5}{4}=0.125$ , and the midpoints are x=[0.5675,0.6925,0.8175,0.9325] .

Since this problem is dealing with the area of the region between functions, it's essentially asking for the difference of the larger and smaller areas. Over the given interval, $e^{\frac{1}{x}}>e^{x^2}$ .

The integral is thus:

${\int_{0.5}^1 (e^{\frac{1}{x}}-e^{x^2})dx \approx 0.125[(e^{\frac{1}{0.5675}}-e^{0.5675^2})+(e^{\frac{1}{0.6925}}-e^{0.6925^2})+(e^{\frac{1}{0.8175}}-e^{0.8175^2})+(e^{\frac{1}{0.9325}}-e^{0.9325^2})]}$

${\int_{0.5}^1 (e^{\frac{1}{x}}-e^{x^2})dx \approx 1.1314$

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Example Question #1082 : Calculus

Using the method of midpoint Reimann sums, approximate the area between the curves $f(x)=e^{\frac{1}{x}}$ and $g(x)=e^{x^2}$ over the interval x=[1,2] using four midpoints.

Possible Answers:

49.7709

36.9908

6.2213

12.4427

24.8854

Correct answer:

12.4427

Explanation:

The Reimann sum approximation of an integral of a function with subintervals over an interval [a,b] takes the form:

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

Where $\frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are four midpoints, the subintervals have length $\frac{2-1}{4}=0.25$ , and the midpoints are x=[1.125,1.375,1.625,1.875] .

The integral is thus:

${\int_1^2 (e^{x^2}-e^{\frac{1}{x}})dx \approx 0.25[(e^{1.125^2}-e^{\frac{1}{1.125}})+(e^{1.375^2}-e^{\frac{1}{1.375}})+(e^{1.625^2}-e^{\frac{1}{1.625}})+(e^{1.875^2}-e^{\frac{1}{1.875}})]}$

${\int_1^2 (e^{x^2}-e^{\frac{1}{x}})dx \approx 12.4427$

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Example Question #1083 : Calculus

The velocity of an errant particle is given by the function $v(t)=\frac{1}{2}e^{tsin(t)}$ . Approximate the average velocity of the particle over the interval of time $t=[0,4\pi]$ using the method of midpoint Reimann sums and four midpoints.

Possible Answers:

1118.3

2136.7

659.2

322.6

4053.9

Correct answer:

322.6

Explanation:

Average velocity can be found as total distance traveled divided by total time elapsed.

Distance traveled over an interval of time can be found by integrating the velocity function over this interval of time. For this problem, Reimann sums will be used to approximate this integral.

The Reimann sum approximation of an integral of a function with subintervals over an interval [a,b] takes the form:

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

Where $\frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are four midpoints, the subintervals have length $\frac{4\pi-0}{4}=\pi$ , and the midpoints are $x=[\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2}]$ .

The integral is thus:

$\int_0^{4\pi} f(x)dx \approx \pi[(\frac{1}{2}e^{\frac{\pi}{2}sin(\frac{\pi}{2})})+(\frac{1}{2}e^{\frac{3\pi}{2}sin(\frac{3\pi}{2})})+(\frac{1}{2}e^{\frac{5\pi}{2}sin(\frac{5\pi}{2})})+(\frac{1}{2}e^{\frac{7\pi}{2}sin(\frac{7\pi}{2})})]$

$\int_0^{4\pi} f(x)dx \approx 4053.9$

Now note that this is an approximation of the distance traveled and we want the average velocity. To find this, divide this value by the time elapsed, which is $4\pi$ :

$v_{avg}=322.6$

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Example Question #1084 : Calculus

Using the method of midpoint Reimann sums, approximate the integral $\int_3^5 x^{\frac{1}{x}}dx$ using five midpoints.

Possible Answers:

10.233

6.281

1.413

7.991

2.827

Correct answer:

2.827

Explanation:

The Reimann sum approximation of an integral of a function with subintervals over an interval [a,b] takes the form:

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

Where $\frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are five midpoints, the subintervals have length $\frac{5-3}{5}=0.4$ , and the midpoints are x=[3.2,3.6,4.0,4.4,4.8] .

This gives the integral approximation:

$\int_3^5 x^{\frac{1}{x}}dx \approx 0.4[3.2^{\frac{1}{3.2}}+3.6^{\frac{1}{3.6}}+4.0^{\frac{1}{4.0}}+4.4^{\frac{1}{4.4}}+4.8^{\frac{1}{4.8}}]$

$\int_3^5 x^{\frac{1}{x}}dx \approx 2.827$

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Example Question #1085 : Calculus

Using the method of midpoint Reimann sums, approximate the integral $\int_0^1 x^{x^x} dx$ using four midpoints.

Possible Answers:

2.301

3.451

0.575

1.150

1.726

Correct answer:

0.575

Explanation:

The Reimann sum approximation of an integral of a function with subintervals over an interval [a,b] takes the form:

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

Where $\frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are four midpoints, the subintervals have length $\frac{1-0}{4}=0.25$ , and the midpoints are x=[0.125,0.375,0.625,0.875] .

This gives the integral approximation:

$\int_0^1 x^{x^x} dx \approx 0.25[0.125^{0.125^{0.125}}+0.375^{0.375^{0.375}}+0.625^{0.625^{0.625}}+0.875^{0.875^{0.875}}]$

$\int_0^1 x^{x^x} dx \approx 0.575$

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Example Question #1086 : Calculus

Approximate the integral $\int_1^4 2^{2^{x}}dx$ using the method of midpoint Reimann sums and three midpoints.

Possible Answers:

276

65808

65842.5

2851.91

2603.01

Correct answer:

2603.01

Explanation:

The Reimann sum approximation of an integral of a function with subintervals over an interval [a,b] takes the form:

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

Where $\frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are three midpoints, the subintervals have length $\frac{4-1}{3}=1$ , and the midpoints are x=[1.5,2.5,3.5] .

Thus this gives the integral approximation:

$\int_1^4 2^{2^{x}}dx \approx [2^{2^{1.5}}+2^{2^{2.5}}+2^{2^{3.5}}]$

$\int_1^4 2^{2^{x}}dx \approx 2603.01$

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Example Question #1087 : Calculus

Approximate the integral $\int_{-\pi}^{\pi}x^{sin(x)}dx$ using the method of midpoint Reimann sums and four midpoints.

Possible Answers:

-2.16i

2.56-2.16i

2.56

2.16i

2.56+2.16i

Correct answer:

2.56-2.16i

Explanation:

The Reimann sum approximation of an integral of a function with subintervals over an interval [a,b] takes the form:

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

Where $\frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are four midpoints, the subintervals have length $\frac{\pi-(-\pi)}{4}=\frac{\pi}{2}$ , and the midpoints are $x=[\frac{-3\pi}{4},\frac{-\pi}{4},\frac{\pi}{4},\frac{3\pi}{4}]$ .

This gives the integral approximation:

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

$\int_{-1}^{1}x^{sin(x)}dx \approx 2.56-2.16i$

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Example Question #1088 : Calculus

Approximate the integral $\int_5^8 x^{\frac{1}{x}}dx$ using the method of midpoint Reimann sums and two midpoints.

Possible Answers:

1.335

4.926

2.670

4.005

1.127

Correct answer:

4.005

Explanation:

The Reimann sum approximation of an integral of a function with subintervals over an interval [a,b] takes the form:

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

Where $\frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are two midpoints, the subintervals have length $\frac{8-5}{2}=1.5$ , and the midpoints are x=[5.75,7.25] .

The integral approximation becomes:

$\int_5^8 x^{\frac{1}{x}}dx \approx 1.5[5.75^{\frac{1}{5.75}}+7.25^{\frac{1}{7.25}}]$

$\int_5^8 x^{\frac{1}{x}}dx \approx 4.005$

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Example Question #1089 : Calculus

Approximate the integral $\int_2^6\frac{1}{x}^{\frac{1}{x}}dx$ using the method of midpoint Reimann sums and two midpoints.

Possible Answers:

2.898

1.299

3.123

1.418

2.836

Correct answer:

2.836

Explanation:

The Reimann sum approximation of an integral of a function with subintervals over an interval [a,b] takes the form:

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

Where $\frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are two midpoints, the subintervals have length $\frac{6-2}{2}=2$ , and the midpoints are x=[3,5] .

This gives the integral approximation

$\int_2^6\frac{1}{x}^{\frac{1}{x}}dx \approx 2[\frac{1}{3}^{\frac{1}{3}}+\frac{1}{5}^{\frac{1}{5}}]$

$\int_2^6\frac{1}{x}^{\frac{1}{x}}dx \approx 2.836$

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Calculus 1 : How to find midpoint Riemann sums

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #1071 : Calculus

Example Question #1081 : Calculus

Example Question #1082 : Calculus

Example Question #1083 : Calculus

Example Question #1084 : Calculus

Example Question #1085 : Calculus

Example Question #1086 : Calculus

Example Question #1087 : Calculus

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