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Calculus 1 : How to find midpoint Riemann sums

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Example Question #101 : How To Find Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{0.25}^{1.75} (3^{x^2}-2^{x^3})dx$ using three midpoints.

Possible Answers:

1.94

1.35

2.03

0.97

2.69

Correct answer:

1.35

Explanation:

A Reimann 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 approximating $\int_{0.25}^{1.75} (3^{x^2}-2^{x^3})dx$

So the interval is [0.25,1.75] , the subintervals have length $\frac{1.75-0.25}{3}=0.5$ , and since we are using the midpoints of each interval, the x-values are [0.5,1,1.5]

$\int_{0.25}^{1.75} (3^{x^2}-2^{x^3})dx \approx 0.5[(3^{0.5^2}-2^{0.5^3})+(3^{1^2}-2^{1^3})+(3^{1.5^2}-2^{1.5^3})]$

$\int_{0.25}^{1.75} (3^{x^2}-2^{x^3})dx \approx 1.35$

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Example Question #102 : How To Find Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{0.95}^{2.15} sin(2^{tan(x)})dx$ using three midpoints.

Possible Answers:

1.02

1.61

-0.93

0.18

-0.37

Correct answer:

-0.37

Explanation:

A Reimann 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 approximating $\int_{0.95}^{2.15} sin(2^{tan(x)})dx$

So the interval is [0.95,2.15] , the subintervals have length $\frac{2.15-0.95}{3}=0.4$ , and since we are using the midpoints of each interval, the x-values are [1.15,1.55,1.95]

$\int_{0.95}^{2.15} sin(2^{tan(x)})dx \approx 0.4[sin(2^{tan(1.15)})+sin(2^{tan(1.55)})+sin(2^{tan(1.95)})]$

$\int_{0.95}^{2.15} sin(2^{tan(x)})dx \approx -0.37$

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Example Question #103 : How To Find Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate $\int_{0.14}^{0.78} tan(e^{-x})dx$ using two midpoints.

Possible Answers:

2.093

0.670

0.484

4.723

1.511

Correct answer:

0.484

Explanation:

A Reimann 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 approximating $\int_{0.14}^{0.78} tan(e^{-x})dx$

So the interval is [0.14,0.78] , the subintervals have length $\frac{0.78-0.14}{2}=0.32$ , and since we are using the midpoints of each interval, the x-values are [0.30,0.62]

$\int_{0.14}^{0.78} tan(e^{-x})dx \approx 0.32[tan(e^{-0.30})+tan(e^{-0.62})]$

$\int_{0.14}^{0.78} tan(e^{-x})dx \approx 0.484$

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Example Question #104 : How To Find Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate the integral $\int_1^2 ln(1.3^{x^{2.4}})$ using four midpoints.

Possible Answers:

0.916

0.735

0.229

1.548

3.096

Correct answer:

0.735

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 approximating $\int_1^2 ln(1.3^{x^{2.4}})$

So the interval is [1,2] , the subintervals have length $\frac{2-1}{4}=0.25$ , and since we are using the midpoints of each interval, the x-values are [1.125,1.375,1.625,1.875]

$\int_1^2 ln(1.3^{x^{2.4}})\approx 0.25[ln(1.3^{1.125^{2.4}})+ln(1.3^{1.375^{2.4}})+ln(1.3^{1.625^{2.4}})+ln(1.3^{1.875^{2.4}})]$

$\int_1^2 ln(1.3^{x^{2.4}})\approx 0.735$

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Example Question #104 : How To Find Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate $\int_{0.03}^{0.96} 4tan(cos(\frac{1}{x}))dx$ using three midpoints.

Possible Answers:

3.043

0.760

0.881

0.404

1.212

Correct answer:

0.760

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 approximating $\int_{0.03}^{0.96} 4tan(cos(\frac{1}{x}))dx$

So the interval is [0.03,0.96] , the subintervals have length $\frac{0.96-0.03}{3}=0.31$ , and since we are using the midpoints of each interval, the x-values are [0.185,0.495,0.805]

$\int_{0.03}^{0.96} 4tan(cos(\frac{1}{x}))dx \approx 0.31[4tan(cos(\frac{1}{0.185}))+4tan(cos(\frac{1}{0.495}))+4tan(cos(\frac{1}{0.805}))]$

$\int_{0.03}^{0.96} 4tan(cos(\frac{1}{x}))dx \approx 0.760$

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Example Question #101 : Midpoint Riemann Sums

Let

f(x)=x

What is the Midpoint Riemann Sum on the interval [0,10] divided into five sub-intervals?

Possible Answers:

Correct answer:

Explanation:

The interval [0,10] divided into four sub-intervals gives rectangles with vertices of the bases at

x=0,2,4,6,8,10

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) , f(7) , and f(9) .

f(1)=1

f(3)=3

f(5)=5

f(7)=7

f(9)=9

Because each sub-interval has a width of , the Midpoint Riemann sum is

A=bh=2(1+3+5+7+9)=50

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Example Question #101 : How To Find Midpoint Riemann Sums

Let

f(x)=x^3-4x^2+4x

What is the Midpoint Riemann Sum on the interval [0,8] divided into four sub-intervals?

Possible Answers:

372

124

200

448

Correct answer:

448

Explanation:

The interval [0,8] divided into four sub-intervals gives rectangles with vertices of the bases at

x=0,2,4,6,8 .

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) .

f(1)=1^3-4*1^2+4*1=1

f(3)=3^3-4*3^2+4*3=3

f(5)=5^3-4*5^2+4*5=45

f(7)=7^3-4*7^2+4*7=175

Because each sub-interval has a width of 2, the Midpoint Riemann sum is

A=bh=2(1+3+45+175)=448 .

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Example Question #108 : How To Find Midpoint Riemann Sums

The general Riemann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

$\frac{b-a}{n}\sum_{i=1}^{n}f(x_i)$

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as I_i ,

the integral approximation becomes

$\sum_{i=1}^{n}I_if(x_i)$

Which of the following parameters would give the closest integral approximation of the function:

$\int_0^{0.5} \frac{1}{2}^{x^2-2x+1}dx$ ?

Possible Answers:

$n=8;I_i>I_{i+1}$

$n=16;I_i<I_{i+1}$

$n=16;I_i>I_{i+1}$

$n=8;I_i<I_{i+1}$

Correct answer:

$n=16;I_i<I_{i+1}$

Explanation:

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=16 will give a closer approximation than n=8:

Reimannsumcomparison

Now, assuming the amount of points/intervals are fixed, but the actual widths can differ, it would be prudent to have narrower intervals at those regions where the graph is steepest. Relatively flatter regions of the function do not lead to the approximation overshooting or undershooting as much as the steeper areas, as can be seen in the above figure.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

Considering $f(x)=\frac{1}{2}^{x^2-2x+1}$ , find the first derivative:

d(a^u)=a^uln(a)du

$f'(x)=(2x-2)\frac{1}{2}^{x^2-2x+1}ln(\frac{1}{2})$

The term in the parenthesis is what may lend to variability in the sign. Over the interval x=[0,0.5] the parenthetical is negative, so the derivative is positive, since $ln(\frac{1}{2})=-0.693$ , and the product of two negatives is positive. Now consider the second derivative:

d(uv)=udv+vdu

$f''(x)=\frac{1}{2}^{x^2-2x+2}ln(\frac{1}{2})+(2x-2)^2\frac{1}{2}^{x^2-2x+1}ln^2(\frac{1}{2})$

The second term will always be positive, and the first term will always be negative. On the interval x=[0,0.5] , it can be found, however, that the second derivative will be positive. If calculating this is to nettlesome, plot this function on the interval:

Save 3

Since the first and second derivatives have opposite signs, the function is getting flatter. The intervals should in turn get wider.

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Example Question #102 : How To Find Midpoint Riemann Sums

The general Riemann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

$\frac{b-a}{n}\sum_{i=1}^{n}f(x_i)$

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as I_i ,

the integral approximation becomes

$\sum_{i=1}^{n}I_if(x_i)$

Which of the following parameters would give the closest integral approximation of the function:

$\int_0^{\frac{\pi}{2}} cos(\frac{x^2}{\pi})dx$ ?

Possible Answers:

$n=8;I_i<I_{i+1}$

$n=16;I_i<I_{i+1}$

$n=8;I_i>I_{i+1}$

$n=16;I_i>I_{i+1}$

Correct answer:

$n=16;I_i>I_{i+1}$

Explanation:

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=16 will give a closer approximation than n=8:

Reimannsumcomparison

Considering $f(x)=cos(\frac{x^2}{\pi})$ take the first derivative:

d(cos(u))=-sin(u)du

$f'(x)=-\frac{2x}{\pi}sin(\frac{x^2}{\pi})$

Within the interval $x=[0,\frac{\pi}{2}]$ this function is negative. Now consider the second derivative:

d(uv)=udv+vdu

$f''(x)=-\frac{2}{\pi}sin(\frac{x^2}{\pi})-\frac{4x^2}{\pi^2}cos(\frac{x^2}{\pi})$

Over the interval $x=[0,\frac{\pi}{2}]$ , the second derivative is negative as well.

Since the signs match, the slope gets steeper, and the intervals should get thinner.

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Example Question #110 : How To Find Midpoint Riemann Sums

The general Riemann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

$\frac{b-a}{n}\sum_{i=1}^{n}f(x_i)$

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as I_i ,

the integral approximation becomes

$\sum_{i=1}^{n}I_if(x_i)$

Which of the following parameters would give the closest integral approximation of the function:

$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} e^{cos(x)}dx$ ?

Possible Answers:

$n=8;I_i<I_{i+1}$

$n=8;I_i>I_{i+1}$

$n=16;I_i>I_{i+1}$

$n=16;I_i<I_{i+1}$

Correct answer:

$n=16;I_i<I_{i+1}$

Explanation:

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=16 will give a closer approximation than n=8:

Reimannsumcomparison

Considering $f(x)= e^{cos(x)}$ , find the first derivative:

d(e^u)=e^udu

d(cos(u))=-sin(u)du

$f'(x)=-sin(x)e^{cos(x)}$

Within the interval $x=[\frac{\pi}{3},\frac{\pi}{2}]$ , the function only takes negative values. Now, take the second derivative:

d(uv)=udv+vdu

$f''(x)=-cos(x)e^{cos(x)}+sin^2(x)e^{cos(x)}$

On the interval $x=[\frac{\pi}{3},\frac{\pi}{2}]$ , this function takes on positive values.

Since the signs of the derivatives are opposing, the function gets flatter over the course of this interval, and the subintervals should thus get wider.

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Calculus 1 : How to find midpoint Riemann sums

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #101 : How To Find Midpoint Riemann Sums

Example Question #102 : How To Find Midpoint Riemann Sums

Example Question #103 : How To Find Midpoint Riemann Sums

Example Question #104 : How To Find Midpoint Riemann Sums

Example Question #104 : How To Find Midpoint Riemann Sums

Example Question #101 : Midpoint Riemann Sums

Example Question #101 : How To Find Midpoint Riemann Sums

Example Question #108 : How To Find Midpoint Riemann Sums

Example Question #102 : How To Find Midpoint Riemann Sums

Example Question #110 : How To Find Midpoint Riemann Sums

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