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Calculus 1 : Functions

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Example Question #81 : Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate the integral $\int_0^{3\pi} cos(sin(cos(x)))dx$ using three midpoints.

Possible Answers:

$3\pi$

$-3\pi$

Correct answer:

$3\pi$

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 .

The integral we're approximating is $\int_0^{3\pi} cos(sin(cos(x)))dx$ .

So the interval is $[0,3\pi]$ , the subintervals have length $\frac{3\pi-0}{3}=\pi$ , and since we are using the midpoints of each interval, the x-values are $\left[\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}\right]$ .

$\int_0^{3\pi} cos(sin(cos(x)))dx \approx \pi [cos(sin(cos(\frac{\pi}{2})))+cos(sin(cos(\frac{3\pi}{2})))+cos(sin(cos(\frac{5\pi}{2})))]$

$\int_0^{3\pi} cos(sin(cos(x)))dx \approx 3\pi$

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Example Question #82 : 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:

$\int_3^{20} (x^2-3x-4)dx$ ?

Possible Answers:

$n=30;I_i>I_{i+1}$

$n=15;I_i>I_{i+1}$

$n=15;I_i<I_{i+1}$

$n=15;I_i\le I_{i+1}$

$n=30;I_i<I_{i+1}$

Correct answer:

$n=30;I_i>I_{i+1}$

Explanation:

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=30 will give a closer approximation than n=15:

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!

The derivative of the function

x^2-3x-4

is found to be

2x-3

Over the interval [3,20] this function, and therefore the slope, is positive.

The second derivative is simpy

meaning that the slope will always be increasing.

Since both the first and second derivatives share the same sign, the function is getting steeper, and the subintervals should get progressively narrower.

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Example Question #83 : 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:

$\int_0^{3} (x^2-8x+16)dx$ ?

Possible Answers:

$n=80;I_i>I_{i+1}$

$n=80;I_i<I_{i+1}$

$n=74;I_i>I_{i+1}$

$n=75;I_i>I_{i+1}$

$n=75;I_i<I_{i+1}$

Correct answer:

$n=80;I_i<I_{i+1}$

Explanation:

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=80 will give a closer approximation than n=75:

Reimannsumcomparison

For the function

x^2-8x+16

The derivative is

2x-8

Over the interval [0,3] this function, and thus the slope, is negative. However, if we take the derivative once more to find the rate of change of the slope, we see that it is

A positive value. Since the first and second derivatives have opposing signs, the function is flattening out. As such, the subintervals should get progressively wider.

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Example Question #84 : 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:

$\int_2^5 (\frac{1}{3}x^3-5x^2-16x+24)dx$ ?

Possible Answers:

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

$n=7;I_i<I_{i+1}$

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

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

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

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 the function

$\frac{1}{3}x^3-5x^2-16x+24$

The derivative is

x^2-10x-16

Which can be rewritten as

(x-2)(x-8)

Over the interval [2,5] the function has negative values. Now consider the second derivative:

2x-10

Over the interval [2,5] , this new function is also negative.

Since the first and second derivatives share signs, the slope is growing steeper. As such, the subintervals should become narrower in accordance.

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Example Question #85 : Midpoint Riemann Sums

The general Reimann 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:

$\int_0^2 (x^3-6x^2+12x+6)dx$ ?

Possible Answers:

$n=50;I_i>I_{i+1}$

$n=40;I_i>I_{i+1}$

$n=40;I_i<I_{i+1}$

$n=50;I_i<I_{i+1}$

Correct answer:

$n=50;I_i<I_{i+1}$

Explanation:

The more points/intervals that are selected, the closer the Reimann sum approximation becomes to the actual integral value, so n=50 will give a closer approximation than n=40:

Reimannsumcomparison

First find the derivative of the function x^3-6x^2+12x+6 :

3x^2-12x+12

This can be rewritten as

3(x-2)(x-2)

Thus for all values for x outside of 2, the slope is positive, though as the function is continuous, a single point is nigh negligible.

Now take the second derivative:

6x-12

For the interval [0,2] , the second derivative is negative.

Since the signs of the first and second derivatives are opposing, the function is flattening out. The intervals should grow wider.

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Example Question #86 : Midpoint Riemann Sums

The general Reimann 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:

$\int_{0.1}^{0.01} x\ln(x)dx$ ?

Possible Answers:

$n=30;I_i>I_{i+1}$

$n=20;I_i<I_{i+1}$

$n=30;I_i<I_{i+1}$

$n=20;I_i>I_{i+1}$

Correct answer:

$n=30;I_i<I_{i+1}$

Explanation:

The more points/intervals that are selected, the closer the Reimann sum approximation becomes to the actual integral value, so n=30 will give a closer approximation than n=20 :

Reimannsumcomparison

For the function

$x\ln(x)$

The derivative can be found, using the product rule d(uv)=udv+vdu :

$\ln(x)+x\frac{1}{x}$

$\ln(x)+1$

Over the entire interval [0.01,0.1] , this function is negative. Now, consider the second derivative:

$\frac{1}{x}$

Over the interval [0.01,0.1] , the function is positive.

Since the first and second derivatives have opposing signs, the function is flattening out. The subintervals should grow progressively thicker.

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Example Question #87 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{\ln(1)}^{\ln(27)} \cos(\pi e^x)dx$ using three midpoints.

Possible Answers:

0.790

0.288

0.137

-0.112

0.565

Correct answer:

0.137

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 .

The integral we're approximating is $\int_{\ln(1)}^{\ln(27)} \cos(\pi e^x)dx$

So the interval is $[\ln(1),\ln(27)]$ , the subintervals have length $\frac{\ln(27)-\ln(1)}{3}=\frac{\ln(27)}{3}$ , and since we are using the midpoints of each interval, the x-values are $[\frac{\ln(27)}{6},\frac{\ln(27)}{2},\frac{5\ln(27)}{6}]$

$\int_{\ln(1)}^{\ln(27)} \cos(\pi e^x)dx \approx \frac{\ln(27)}{3}[\cos(\pi e^{\frac{\ln(27)}{6}})+\cos(\pi e^{\frac{\ln(27)}{2}})+\cos(\pi e^{\frac{5\ln(27)}{6}})]$

$\int_{\ln(1)}^{\ln(27)} \cos(\pi e^x)dx \approx 0.137$

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Example Question #88 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{\ln(3)}^{\ln(81)}e^{2x}dx$ using three midpoints.

Possible Answers:

2457

819

$2457\ln(3)$

$819\ln(3)$

Correct answer:

$2457\ln(3)$

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 asked to approximate $\int_{\ln(3)}^{\ln(81)}e^{2x}dx$

So the interval is $[\ln(3),\ln(81)]$ , the subintervals have length $\frac{\ln(81)-\ln(3)}{3}=\frac{\ln(27)}{3}=\ln(3)$ , and since we are using the midpoints of each interval, the x-values are $[\frac{3\ln(3)}{2},\frac{5\ln(3)}{2},\frac{7\ln(3)}{2}]$

$\int_{\ln(3)}^{\ln(81)}e^{2x}dx \approx \ln(3)[e^{2(\frac{3\ln(3)}{2})}+e^{2(\frac{5\ln(3)}{2})}+e^{2(\frac{7\ln(3)}{2})}]$

$\int_{\ln(3)}^{\ln(81)}e^{2x}dx \approx 2457\ln(3)$

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Example Question #89 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{\pi}^{4\pi} e^{\cos(\ln(x))}dx$ using three midpoints.

Possible Answers:

6.67

2.12

1.22

4.89

9.54

Correct answer:

6.67

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 $\int_{\pi}^{4\pi} e^{\cos(\ln(x))}dx$

So the interval is $[\pi,4\pi]$ , the subintervals have length $\frac{4\pi-\pi}{3}=\pi$ , and since we are using the midpoints of each interval, the x-values are $[\frac{3}{2}\pi,\frac{5}{2}\pi,\frac{7}{2}\pi]$

$\int_{\pi}^{4\pi} e^{\cos(\ln(x))}dx \approx \pi[e^{\cos(\ln(\frac{3\pi}{2}))}+e^{\cos(\ln(\frac{5\pi}{2}))}+e^{\cos(\ln(\frac{7\pi}{2}))}]$

$\int_{\pi}^{4\pi} e^{\cos(\ln(x))}dx \approx 6.67$

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Example Question #90 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{-5}^{-2}x^{e^x}dx$ using three midpoints

Possible Answers:

3.09 + 0.41i

0.41-3.09i

3.09 - 0.41i

0.41+3.09i

Correct answer:

3.09 + 0.41i

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

So the interval is [-5,-2] , the subintervals have length $\frac{-2-(-5)}{3}=1$ , and since we are using the midpoints of each interval, the x-values are [-4.5,-3.5,-2.5]

$\int_{-5}^{-2}x^{e^x}dx \approx 1[(-4.5)^{e^{(-4.5)}}+(-3.5)^{e^{(-3.5)}}+(-2.5)^{e^{(-2.5)}}]$

$\int_{-5}^{-2}x^{e^x}dx \approx 3.09 + 0.41i$

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #81 : Midpoint Riemann Sums

Example Question #82 : Midpoint Riemann Sums

Example Question #83 : Midpoint Riemann Sums

Example Question #84 : Midpoint Riemann Sums

Example Question #85 : Midpoint Riemann Sums

Example Question #86 : Midpoint Riemann Sums

Example Question #87 : Midpoint Riemann Sums

Example Question #88 : Midpoint Riemann Sums

Example Question #89 : Midpoint Riemann Sums

Example Question #90 : Midpoint Riemann Sums

All Calculus 1 Resources

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