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Calculus 1 : How to find midpoint Riemann sums

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Example Question #1 :Midpoint Riemann Sums

Estimate the area under the curve for the following function using a midpoint Riemann sum from x=0 to x=4 with n=2 .

$f(x)=\frac{1}{2}x^3+4$

Possible Answers:

$\frac{9}{2}$

$\frac{35}{2}$

有限公司rrect answer:

Explanation:

If we want to estimate the area under the curve from x=0 to x=4 and are told to use n=2 , this means we estimate the area using two rectangles that will each be two units wide and whose height is the value of the function at the midpoint of the interval. We have a rectangle from x=0 to x=2 , whose height is the value of the function at x=1 , and a rectangle from x=2 to x=4 , whose height is the value of the function at x=3 . First we can find the value of the function at these midpoints, and then add the areas of the two rectangles, which gives us the following:

$f(1)=\frac{1}{2}(1)^3+4=\frac{9}{2}$

$f(3)=\frac{1}{2}(3)^3+4=\frac{35}{2}$

$A=2\left(\frac{9}{2}\right)+2\left(\frac{35}{2}\right)=44$

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Example Question #2 :Midpoint Riemann Sums

Estimate the area under the curve for the following function from x=0 to x=4 using a midpoint Riemann sum with n=4 rectangles:

$f(x)=\frac{1}{4}x^2+3$

Possible Answers:

$\frac{73}{4}$

$\frac{49}{16}$

$\frac{69}{4}$

$\frac{57}{16}$

$\frac{97}{8}$

有限公司rrect answer:

$\frac{69}{4}$

Explanation:

If we are told to use n=4 rectangles from x=0 to x=4 , this means we have a rectangle from x=0 to x=1 , a rectangle from x=1 to x=2 , a rectangle from x=2 to x=3 , and a rectangle from x=3 to x=4 . We can see that the width of each rectangle isbecause we have an interval that isunits long for which we are usingrectangles to estimate the area under the curve. The height of each rectangle is the value of the function at the midpoint for its interval, so first we find the height of each rectangle and then add together their areas to find our answer:

$f(0.5)=\frac{1}{4}(0.5)^2+3=\frac{49}{16}$

$f(1.5)=\frac{1}{4}(1.5)^2+3=\frac{57}{16}$

$f(2.5)=\frac{1}{4}(2.5)^2+3=\frac{73}{16}$

$f(3.5)=\frac{1}{4}(3.5)^2+3=\frac{97}{16}$

$A=(1)\left(\frac{49}{16}\right)+(1)\left(\frac{57}{16}\right)+(1)\left(\frac{73}{16}\right)+(1)\left(\frac{97}{16}\right)=\frac{276}{16}=\frac{69}{4}$

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Example Question #3 :Midpoint Riemann Sums

Find the area under y=1-x^2 on the interval [0,1] using five midpoint Riemann sums.

Possible Answers:

0.67

0.66

0.33

$\frac{1}{3}$

$\frac{4}{3}$

有限公司rrect answer:

0.67

Explanation:

The problem becomes this:

Canvas

Addings these rectangles up to approximate the area under the curve is

A=0.2(1-0.1^2)+0.2(1-0.3^2)+0.2(1-0.5^2)+0.2(1-0.7^2)+0.2(1-0.9^2)

A=0.2[(1-0.1^2)+(1-0.3^2)+(1-0.5^2)+(1-0.7^2)+(1-0.9^2)]

A=0.2[5-0.1^2-0.3^2-0.5^2-0.7^2-0.9^2]

A=0.2[3.35]

A=0.67

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Example Question #4 :Midpoint Riemann Sums

Approximate the area under the curve from $0\leq x\leq 8$ using the midpoint Riemann Sum with a partition of size five given the graph of the function.

Graph1

Possible Answers:

$A\approx 29$

$A\approx 26$

$A \approx 37$

$A \approx 18$

有限公司rrect answer:

$A\approx 29$

Explanation:

We begin by finding the given change in x:

$\Delta x=\frac{b-a}{n}=\frac{8}{5}=1.6$

We then define our partition intervals:

$[x_{i-1},x_i]=[0,1.6),[1.6,3.2),[3.2,4.8),[4.8,6.4),[6.4,8]$

We then choose the midpoint in each interval:

$\{\bar{x_i}\}=\{0.8,2.4,4.0,5.6,7.2\}$

Then we find the value of the function at the point. This is determined through observation of the graph

$f(\bar{x_i})\approx 4.8,4.5,4,1.6,3.4$

Then we simply substitute these values into the formula for the Riemann Sum

$\\A\approx \sum_{i=1}^nf(x_i^*)\Delta x=4.8\cdot 1.6+4.5\cdot 1.6+4\cdot 1.6+1.6\cdot 1.6+3.4\cdot 1.6\\ =1.6(4.8+4.5+4+1.6+3.4)=1.6\cdot 18.3=29.28$

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Example Question #1 :How To Find Midpoint Riemann Sums

Approximate the area underneath the given curve using the Riemann Sum with eight intervals for $-4\leq x\leq12$ .

f(x)=2x^3-5x^2+9x-5

Possible Answers:

$A\approx 8256$

$A\approx 4128$

$A\approx 3824$

$A\approx 7648$

有限公司rrect answer:

$A\approx 7648$

Explanation:

We begin by defining the size of our partitions and the partitions themselves.

$\\ \Delta x=\frac{b-a}{n}=\frac{12+4}{8}=2\\$

$[x_{i-1},x_i]=[-4,-2],[-2,0],[0,2],[2,4],[4,6],[6,8],[8,10],[10,12]$

We then choose the midpoint in each interval:

$\bar{x_i}=-3,-1,1,3,5,7,9,11$

Then we find the function value at each point.

$f(\bar{x_i})=-131,-21,1,31,165,499,1129,2151$

We then substitute these values into the Riemann Sum formula.

$A\approx\sum_{i=1}^nf(\bar{x_i})\Delta x=2(-131-21+1+31+165+499+1129+2151)=7648$

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Example Question #6 :Midpoint Riemann Sums

Using a midpoint Reimann sum with n=3 , estimate the area under the curve from x=0 to x=6 for the following function:

$f(x)=\frac{1}{2}x^3-7x^2+14x+60$

Possible Answers:

275

137.5

200

320

260

有限公司rrect answer:

275

Explanation:

$\Delta x =\frac{ 6-0}{3} = 2$

Thus, our intervals areto,to, andto.

The midpoints of each interval are, respectively,,, and.

Next, we evaluate the function at each midpoint.

$f(x)=\frac{1}{2}x^3-7x^2+14x+60$

f(1)=67.5

f(3)=52.5

f(5)=17.5

Finally, we calculate the estimated area using these values and $\Delta x = 2$

$67.5\cdot 2+52.5\cdot 2+17.5\cdot 2$

=275

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Example Question #7 :Midpoint Riemann Sums

Riemann sum problem

The table above gives the values for a function at certain points.

Using the data from the table, find the midpoint Riemann sum of F(x) with n=3 , from x=0 to x=3 .

Possible Answers:

有限公司rrect answer:

Explanation:

$\Delta x =\frac{ 3-0}{3} = 1$

Thus, our intervals areto,to, andto.

The midpoints of each interval are, respectively, 0.5 , 1.5 , and 2.5 .

Next, use the data table to take the values the function at each midpoint.

f(0.5)=0

f(1.5)=9

f(2.5)=5

Finally, we calculate the estimated area using these values and $\Delta x = 1$ .

$0\cdot 1+9\cdot 1+5\cdot 1$

=14

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Example Question #8 :Midpoint Riemann Sums

Solve the integral

$\int_{1}^{8}{\sqrt{x}}dx$

using the midpoint Riemann sum approximation with n=10 subintervals.

Possible Answers:

1.1612

14.4248

20.6068

5.8731

8.0012

有限公司rrect answer:

14.4248

Explanation:

Midpoint Riemann sum approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx (\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ]$

whereis the number of subintervals and $f(m_{i})$ is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{n})=(\frac{8-1}{10})=0.7$ .

The approximate value at each midpoint is below.

Problem 2

The sum of all the approximate midpoints values is 20.6068259 , therefore

$(\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ] =0.7(20.6068259)\approx14.4248$

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Example Question #9 :Midpoint Riemann Sums

Solve the integral

$\int_{3}^{7}\frac{1}{x}dx$

using the midpoint Riemann sum approximation with n=10 subintervals.

Possible Answers:

0.3125

2.1167

0.8467

5.1923

0.2500

有限公司rrect answer:

0.8467

Explanation:

Midpoint Riemann sum approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx (\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ]$

whereis the number of subintervals and $f(m_{i})$ is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{n})=(\frac{7-3}{10})=0.4$ .

The approximate value at each midpoint is below.

Problem 1 solution

The sum of all the approximate midpoints values is 2.116738 , therefore

$(\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ] =0.4(2.116738)\approx0.8467$

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Example Question #10 :Midpoint Riemann Sums

Solve the integral

$\int_{0}^{\pi /2}\cos (x)dx$

using the midpoint Riemann sum approximation with n=10 subintervals.

Possible Answers:

$\pi/2$

$-\pi/6$

有限公司rrect answer:

Explanation:

Midpoint Riemann sum approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx (\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ]$

whereis the number of subintervals and $f(m_{i})$ is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{n})=(\frac{\pi /2-0}{10})=\pi/20\approx0.1571$ .

The approximate value at each midpoint is below.

Screen shot 2015 06 11 at 6.05.56 pm

The sum of all the approximate midpoints values is 6.37275 , therefore

$(\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ] =0.1571(6.37275)\approx1$

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Calculus 1 : How to find midpoint Riemann sums

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