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Calculus 1 : Differential Functions

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Example Question #1081 : Calculus

Using the method of midpoint Reimann sums, approximate the integral $\int_0^{10\pi} xcos(\frac{1}{\pi^2}x^3)dx$ using five midpoints.

Possible Answers:

$50\pi^2$

$-50\pi^2$

$-50\pi$

$50\pi^2$

Correct answer:

$-50\pi^2$

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 are given the function

$\int_0^{10\pi} xcos(\frac{1}{\pi^2}x^3)dx$

So the interval is $[0,10\pi]$ , the subintervals have length $\frac{10\pi-0}{5}=2\pi$ , and since we are using the midpoints of each interval, the x-values are $[\pi,3\pi,5\pi,7\pi,9\pi]$

The Reimann sum approximation is thus:

$\int_0^{10\pi} xcos(\frac{1}{\pi^2}x^3)dx \approx 2\pi\sum_{i=1}^{5}(2i-1)\pi cos(\frac{1}{\pi^2}[(2i-1)\pi]^3)$

Or, in long form:

$\int_0^{10\pi} xcos(\frac{1}{\pi^2}x^3)dx \approx {2\pi(\pi cos(\frac{1}{\pi^2}\pi^3)+3\pi cos(\frac{1}{\pi^2}(3\pi)^3)+5\pi cos(\frac{1}{\pi}^2}(5\pi)^3)+7\pi cos(\frac{1}{\pi^2}(7\pi)^3)+9\pi cos(\frac{1}{\pi^2}(9\pi)^3))}$

Note that $cos(n\pi)=-1;(n:odd -integer)$

$\int_0^{10\pi} xcos(\frac{1}{\pi^2}x^3)dx \approx 2\pi(-\pi-3\pi-5\pi-7\pi-9\pi)$

$\int_0^{10\pi} xcos(\frac{1}{\pi^2}x^3)dx \approx-50\pi^2$

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Example Question #61 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{\pi}^{4\pi} \frac{sin(x)}{x^2}dx$ using three midpoints

Possible Answers:

$\frac{836}{225}\pi$

$\frac{836}{75\pi}$

$\frac{836}{225\pi}$

$\frac{836}{75\pi^2}$

$\frac{836}{225\pi^2}$

Correct answer:

$\frac{836}{225\pi}$

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 .

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 $[0.5\pi,1.5\pi,2.5\pi]$

$\int_{\pi}^{4\pi} \frac{sin(x)}{x^2}dx \approx \pi[\frac{sin(0.5\pi)}{(0.5\pi)^2}+\frac{sin(1.5\pi)}{(1.5\pi)^2}+\frac{sin(2.5\pi)}{(2.5\pi)^2}]$

$\int_{\pi}^{4\pi} \frac{sin(x)}{x^2}dx \approx \frac{836}{225\pi}$

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Example Question #62 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_5^{11} \frac{ln^2(x)}{x^3}$ using three midpoints.

Possible Answers:

0.0572

0.0286

Correct answer:

0.0572

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 .

Our integral is $\int_5^{11} \frac{ln^2(x)}{x^3}$

So the interval is [5,11] , the subintervals have length $\frac{11-5}{3}=2$ , and since we are using the midpoints of each interval, the x-values are [6,8,10]

$\int_5^{11} \frac{ln^2(x)}{x^3} \approx 2[\frac{ln^2(6)}{6^3}+\frac{ln^2(8)}{8^3}+\frac{ln^2(10)}{10^3}]$

$\int_5^{11} \frac{ln^2(x)}{x^3} \approx 0.0572$

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Example Question #63 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{10\pi}^{16\pi}\frac{cos(x^2)}{x}dx$ using three midpoints.

Possible Answers:

-0.235

-0.469

0.469

-0.104

-0.052

Correct answer:

-0.104

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 function given is $\int_{10}^{16}\frac{cos(x^2)}{x}dx$

So the interval is $[10\pi,16\pi]$ , the subintervals have length $\frac{16\pi-10\pi}{3}=2\pi$ , and since we are using the midpoints of each interval, the x-values are $[11\pi,13\pi,15\pi]$

$\int_{10}^{16}\frac{cos(x^2)}{x}dx \approx 2\pi[\frac{cos((11\pi)^2)}{11\pi}+\frac{cos((13\pi)^2)}{13\pi}+\frac{cos((15\pi)^2)}{15\pi}]$

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Example Question #64 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_0^6 (x^2-x)dx$ using three midpoints.

Possible Answers:

Correct answer:

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 .

Our function is $\int_0^6 (x^2-x)dx$

So the interval is [0,6] , the subintervals have length $\frac{6-0}{3}=2$ , and since we are using the midpoints of each interval, the x-values are [1,3,5]

$\int_0^6 (x^2-x)dx\approx 2[(1^2-1)+(3^2-3)+(5^2-5)]$

$\int_0^6 (x^2-x)dx \approx 52$

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Example Question #65 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_0^{12} \frac{x+4}{x+8}dx$ using three midpoints.

Possible Answers:

$\frac{2636}{945}$

$\frac{2636}{315}$

$\frac{659}{945}$

$\frac{10544}{315}$

$\frac{10544}{945}$

Correct answer:

$\frac{2636}{315}$

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 .

Our integral is $\int_0^{12} \frac{x+4}{x+8}dx$

So the interval is [0,12] , the subintervals have length $\frac{12-0}{3}=4$ , and since we are using the midpoints of each interval, the x-values are [2,6,10]

$\int_0^{12} \frac{x+4}{x+8}dx \approx 4[\frac{2+4}{2+8}+\frac{6+4}{6+8}+\frac{10+4}{10+8}]$

$\int_0^{12} \frac{x+4}{x+8}dx \approx \frac{2636}{315}$

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Example Question #66 : Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate the integral $\int_1^4 x^{x-1}dx$ using three midpoints.

Possible Answers:

12.0

28.1

75.0

14.0

37.3

Correct answer:

28.1

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 are given is $\int_1^4 x^{x-1}dx$

So the interval is [1,4] , the subintervals have length $\frac{4-1}{3}=1$ , and since we are using the midpoints of each interval, the x-values are [1.5,2.5,3.5]

$\int_1^4 x^{x-1}dx \approx 1(1.5^{1.5-1}+2.5^{2.5-1}+3.5^{3.5-1})$

$\int_1^4 x^{x-1}dx \approx (1.5^{0.5}+2.5^{1.5}+3.5^{2.5})$

$\int_1^4 x^{x-1}dx \approx 28.1$

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Example Question #67 : Midpoint Riemann Sums

The method of Riemann sums is a way of approximating integrals, often for functions which do not have a defined integral.

Picture an integral $\int_a^b f(x)dx$

A number of rectangular areas are defined, this number being defined by the amount of points or subintervals selected. Generally, these rectangles have uniform widths (that of the subinterval), given by the formula $\frac{b-a}{n}$ where n is the number of points/subintervals.

Each rectangle differs in height, however, with individual heights defined by the function value at a given point, f(x_i) .

The area of a particular rectangle is then $\frac{b-a}{n}f(x)dx$

Adding each of these rectangles together then gives the integral approximation.

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

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:

$\int_0^4 x^xdx$ ?

Possible Answers:

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

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

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

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

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

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.

Since the function x^x becomes progressively steeper, a better approximation for a fixed number of intervals would have the intervals grow thinner as increases.

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Example Question #61 : Functions

Using the method of midpoint Riemann sums, approximate the integral $\int_{10\pi}^{16\pi}e^{\frac{cos(x)}{x}}dx$ using three midpoints.

Possible Answers:

9.19

18.39

3.67

5.85

2.93

Correct answer:

18.39

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 given is $\int_{10\pi}^{16\pi}e^{\frac{cos(x)}{x}}dx$

So the interval is $[10\pi,16\pi]$ , the subintervals have length $\frac{16\pi-10\pi}{3}=2\pi$ , and since we are using the midpoints of each interval, the x-values are $[11\pi,13\pi,15\pi]$

$\int_{10\pi}^{16\pi}e^{\frac{cos(x)}{x}}dx \approx 2\pi[e^{\frac{cos(11\pi)}{11\pi}}+e^{\frac{cos(13\pi)}{13\pi}}}+e^{\frac{cos(15\pi)}{15\pi}}}]$

$\int_{10\pi}^{16\pi}e^{\frac{cos(x)}{x}}dx \approx 18.39$

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Example Question #69 : Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate the area of the region between the functions $f(x)=x^{2x}$ and g(x)=x^x over the interval x=[1,2.5] using three midpoints.

Possible Answers:

37.10

9.27

11.23

19.28

18.55

Correct answer:

18.55

Explanation:

To find the area between two functions, the method traditionally used is to find the difference of the integral values between the larger and smaller function over the specified interval. However, neither of these functions can be integrated, so a method of approximation is useful.

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 told that we're integrating from 1 to 2.5, so the interval is [1,2.5] , the subintervals have length $\frac{2.5-1}{3}=0.5$ , and since we are using the midpoints of each interval, the x-values are [1.25,1.75,2.25] .

As $f(x)=x^{2x}$ is greater than g(x)=x^x over the given interval, our approximation looks as follows:

${\int_{1}^{2.5}(x^{2x}-x^x)dx \approx 0.5[(1.25^{2(1.25)}-1.25^{1.25})+(1.75^{2(1.75)}-1.75^{1.75})+(2.25^{2(2.25)}-2.25^{2.25})]}$

${\int_{1}^{2.5}(x^{2x}-x^x)dx \approx 18.55$

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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1081 : Calculus

Example Question #61 : Midpoint Riemann Sums

Example Question #62 : Midpoint Riemann Sums

Example Question #63 : Midpoint Riemann Sums

Example Question #64 : Midpoint Riemann Sums

Example Question #65 : Midpoint Riemann Sums

Example Question #66 : Midpoint Riemann Sums

Example Question #67 : Midpoint Riemann Sums

Example Question #61 : Functions

Example Question #69 : Midpoint Riemann Sums

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