Midpoint Riemann Sums - CLEP Calculus

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Question

Approximate the $\int_{1}^{9}\frac{10}{x^2+2}dx$ using midpoint Riemann sum when .

Answer

The find the area under the curve for the function you use the formula $\int_{a}^{b}f(x)dx$ .

Riemann's midpoint formula states that

$\int_{a}^{b}f(x)dx= \frac{b-a}{n}(f(x_{1})+f(x_{2})+f(x_{3})+...+f(x_{n}))$ .

This method breaks the area under the curve into n rectangles. The first term of the equation that is multiplied out find the base length of each rectangle and then the f(x) terms are the heights at the middle of each rectangle. By multiplying the two terms, you find the area and when you add them together you find the approximation of the area under the curve.

We will plug in our values into this formula. a=1, b=9 and n=4. To find the x values to plug into the function we look at the base length of each rectangle.

$\frac{b-a}{n}=\frac{9-1}{4}=2$

so there is a rectangle wall every x=2 starting at x=1, thus the midpoints will be x1=2, x2=4, x3=6, x4=8.

$\ \int_{a}^{b}f(x)dx= \frac{9-1}{4}(f(2)+f(4)+f(6)+f(8))\ = 2\left(\frac{5}{3}+\frac{5}{9}+\frac{5}{19}+\frac{5}{33}\right)\ \approx 5.28$

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Question

Approximate the $\int_{1}^{3}\frac{4x+3}{2}dx$ using midpoint Riemann sum when .

Answer

The find the area under the curve for the function you use the formula $\int_{a}^{b}f(x)dx$ .

Riemann's midpoint formula states that

$\int_{a}^{b}f(x)dx= \frac{b-a}{n}(f(x_{1})+f(x_{2})+f(x_{3})+...+f(x_{n}))$ .

This method breaks the area under the curve into n rectangles. The first term of the equation that is multiplied out find the base length of each rectangle and then the f(x) terms are the heights at the middle of each rectangle. By multiplying the two terms, you find the area and when you add them together you find the approximation of the area under the curve.

We will plug in our values into this formula. a=1, b=3 and n=4. To find the x values to plug into the function we look at the base length of each rectangle.

$\frac{b-a}{n}=\frac{3-1}{4}=0.5$

So there is a rectangle wall every x=0.5 starting at x=1, thus the midpoints will be x1=1.25, x2=1.75, x3=2.25, x4=2.75.

$\ \int_{a}^{b}f(x)dx= \frac{3-1}{4}(f(1.25)+f(1.75)+f(2.25)+f(2.75))\ = 0.5(4+5+6+7)\=11$

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Question

Let

What is the Midpoint Riemann Sum on the interval divided into four sub-intervals?

Answer

The interval divided into four sub-intervals gives rectangles with vertices of the bases at

For the Midpoint Riemann sum, we need to find the rectangle heights which values come from the midpoint of the sub-intervals, or f(1), f(3), f(5), and f(7).

Because each interval has width 2, the approximated Midpoint Riemann Sum is

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Question

Estimate the area under the curve for the following function using a midpoint Riemann sum from to with .

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

Answer

If we want to estimate the area under the curve from to and are told to use , 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 to , whose height is the value of the function at , and a rectangle from to , whose height is the value of the function at . 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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Question

Estimate the area under the curve for the following function from to using a midpoint Riemann sum with rectangles:

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

Answer

If we are told to use rectangles from to , this means we have a rectangle from to , a rectangle from to , a rectangle from to , and a rectangle from to . We can see that the width of each rectangle is because we have an interval that is units long for which we are using rectangles 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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Question

Find the area under on the interval using five midpoint Riemann sums.

Answer

The problem becomes this:

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

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Question

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 using four midpoints.

Answer

The Reimann sum approximation of an integral of a function with subintervals over an interval 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 .

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:

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

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

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Question

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.

Answer

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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Question

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

Answer

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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Question

Solve the integral

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

using the midpoint Riemann sum approximation with subintervals.

Answer

Midpoint Riemann sum approximations are solved using the formula

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

where is 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.

The sum of all the approximate midpoints values is , 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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Question

Solve the integral

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

using the midpoint Riemann sum approximation with subintervals.

Answer

Midpoint Riemann sum approximations are solved using the formula

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

where is 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.

The sum of all the approximate midpoints values is , 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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Question

Solve the integral

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

using the midpoint Riemann sum approximation with subintervals.

Answer

Midpoint Riemann sum approximations are solved using the formula

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

where is 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.

The sum of all the approximate midpoints values is , 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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Question

Solve the integral

$\int_{2}^{5}{e{^{x}}}dx$

using the midpoint Riemann sum approximation with subintervals.

Answer

Midpoint Riemann sum approximations are solved using the formula

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

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

For this problem, $(\frac{b-a}{n})=(\frac{5-2}{10})=0.3$ .

The approximate value at each midpoint is below.

The sum of all the approximate midpoints values is , therefore

$(\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ] =0.3(468.3221)\approx140.4966$

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Question

Solve the integral

$\int_{2}^{7}{\frac{1}{5x+3}}dx$

using the trapezoidal approximation with subintervals.

Answer

Trapezoidal approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx T_{n}=(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]$

where is the number of subintervals and is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{7-2}{25})=0.5$ .

The value of each approximation term is below.

The sum of all the approximation terms is , therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.5)(0.86027754)=0.4301$

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Question

Solve the integral

$\int_{1}^{2}{x\sin (x)}dx$

using the trapezoidal approximation with subintervals.

Answer

Trapezoidal approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx T_{n}=(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]$

where is the number of subintervals and is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{2-1}{25})=0.1$ .

The value of each approximation term is below.

The sum of all the approximation terms is , therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.1)(28.786699)=2.8787$

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Question

Solve the integral

$\int_{0}^{4}{\sqrt{5+x^{^{3}}}}dx$

using the trapezoidal approximation with subintervals.

Answer

Trapezoidal approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx =T_{n}=(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]$

where is the number of subintervals and is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{4-0}{25})=0.4$ .

The value of each approximation term is below.

The sum of all the approximation terms is , therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.4)(83.89553)=33.5582$

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Question

Solve the integral

$\int_{1}^{2}{\sqrt{ln(x)}}dx$

using the trapezoidal approximation with subintervals.

Answer

Trapezoidal approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx T_{n}=(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]$

where is the number of subintervals and is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{2-1}{25})=0.1$ .

The value of each approximation term is below.

The sum of all the approximation terms is , therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.1)(11.7257431)=1.1726$

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Question

Solve the integral

$\int_{1}^{5}{\frac{1}{x-7}}dx$

using Simpson's rule with subintervals.

Answer

Simpson's rule is solved using the formula

$\int_{a}^{b}f(x))dx =S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]$

where is the number of subintervals and is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{5-1}{10})=0.4$ .

The value of each approximation term is below.

The sum of all the approximation terms is therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.4)*(-8.23995)=-1.0987$

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Question

Solve the integral

$\int_{8}^{15}{ln(x)}dx$

using Simpson's rule with subintervals.

Answer

Simpson's rule is solved using the formula

$\int_{a}^{b}f(x))dx=S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]$

where is the number of subintervals and is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{15-8}{10})=0.7$ .

The value of each approximation term is below.

The sum of all the approximation terms is therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.7)*(72.79378)=16.9852$

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Question

Solve the integral

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

using Simpson's rule with subintervals.

Answer

Simpson's rule is solved using the formula

$\int_{a}^{b}f(x))dx =S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]$

where is the number of subintervals and is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{1+1}{10})=0.2$ .

The value of each approximation term is below.

The sum of all the approximation terms is therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.2)*(10.5840)=0.7056$

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