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Calculus 1 : Calculus

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #172 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate the average of $f(x)=\frac{tan(x)}{x^3}$ over the interval (1,9) using four midpoints.

Possible Answers:

1.024

0.027

0.386

-0.067

-0.510

Correct answer:

-0.067

Explanation:

To find the average of a function over a given interval of values [a,b] , the most precise method is to use an integral as follows:

$\frac{1}{b-a}\int_{a}^{b} f(x)dx$

Now for functions that are difficult or impossible to integrate,a Riemann sum can be used to approximate the value. 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 equal to the subinterval length $\frac{b-a}{n}$ , and variable heights f(x_i) , which depend on the function value at a given point x_i .

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

$\frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

$\frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))$

We're asked to approximate the average of $f(x)=\frac{tan(x)}{x^3}$ over the interval (1,9)

The subintervals have length $\frac{9-1}{4}=2$ , and since we are using the midpoints of each interval, the x-values are [2,4,6,8]

$\frac{1}{8}\int_{1}^{9}\frac{tan(x)}{x^3}dx\approx (\frac{1}{4})[\frac{tan(2)}{2^3}+\frac{tan(4)}{4^3}+\frac{tan(6)}{6^3}+\frac{tan(8)}{8^3}]$

$\frac{1}{8}\int_{1}^{9}\frac{tan(x)}{x^3}dx\approx -0.067$

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Example Question #173 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{1.5}^{4.5}tan^3(x^3)dx$ using three midpoints.

Possible Answers:

-336.53

45.99

402.56

188.13

-229.01

Correct answer:

-336.53

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 asked to approximate $\int_{1.5}^{4.5}tan^3(x^3)dx$

So the interval is [1.5,4.5] , the subintervals have length $\frac{4.5-1.5}{3}=1$ , and since we are using the midpoints of each interval, the x-values are [2,3,4]

$\int_{1.5}^{4.5}tan^3(x^3)dx\approx (1)[tan^3(2^3)+tan^3(3^3)+tan^3(4^3)]$

$\int_{1.5}^{4.5}tan^3(x^3)dx\approx -336.53$

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Example Question #174 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{1.25}^{2.75}1.65^{x^{4.2}}dx$ using threemidpoints.

Possible Answers:

$8 \cdot 10^7$

$4 \cdot 10^9$

$4 \cdot 10^8$

$8 \cdot 10^9$

$8 \cdot 10^8$

Correct answer:

$8 \cdot 10^9$

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 asked to approximate $\int_{1.25}^{2.75}1.65^{x^{4.2}}dx$

So the interval is [1.25,2.75] , the subintervals have length $\frac{2.75-1.25}{3}=0.5$ , and since we are using the midpoints of each interval, the x-values are [1.5,2,2.5]

$\int_{1.25}^{2.75}1.65^{x^{4.2}}dx\approx (0.5)[1.65^{1.5^{4.2}}+1.65^{2^{4.2}}+1.65^{2.5^{4.2}}]$

$\int_{1.25}^{2.75}1.65^{x^{4.2}}dx\approx 8 \cdot 10^9$

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Example Question #175 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{0.25}^{1.75}sin^2(x)tan^2(x^2)dx$ using three midpoints.

Possible Answers:

-218.14

-5.08

42.09

1.63

24.56

Correct answer:

1.63

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 asked to approximate $\int_{0.25}^{1.75}sin^2(x)tan^2(x^2)dx$

So the interval is [.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]

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

$\int_{0.25}^{1.75}sin^2(x)tan^2(x^2)dx\approx 1.63$

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Example Question #176 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{2.125}^{2.875}cos^3(x^3\pi)dx$ using three midpoints.

Possible Answers:

0.82

0.29

-0.13

0.43

-0.04

Correct answer:

-0.13

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 asked to approximate $\int_{2.125}^{2.875}cos^3(x^3\pi)dx$

So the interval is [2.125,2.875] , the subintervals have length $\frac{2.875-2.125}{3}=0.25$ , and since we are using the midpoints of each interval, the x-values are [2.25,2.50,2.75]

$\int_{2.125}^{2.875}cos^3(x^3\pi)dx\approx (0.25)[cos^3(2.25^3\pi)+cos^3(2.5^3\pi)+cos^3(2.75^3\pi)]$

$\int_{2.125}^{2.875}cos^3(x^3\pi)dx\approx -0.13$

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Example Question #177 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{1.5}^{10.5}cos^4(\pi x^8)dx$ using three midpoints.

Possible Answers:

Correct answer:

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 asked to approximate $\int_{1.5}^{10.5}cos^4(\pi x^8)dx$

So the interval is [1.5,10.5] , the subintervals have length $\frac{10.5-1.5}{3}=3$ , and since we are using the midpoints of each interval, the x-values are [3,6,9]

$\int_{1.5}^{10.5}cos^4(\pi x^8)dx\approx (3)[cos^4(\pi (3)^8)+cos^4(\pi (6)^8)+cos^4(\pi (9)^8)]$

$\int_{1.5}^{10.5}cos^4(\pi x^8)dx\approx 9$

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Example Question #178 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{0.05}^{0.35}14^{2\sqrt{x}}dx$ using three midpoints.

Possible Answers:

33.91

5.02

3.39

1.19

50.23

Correct answer:

3.39

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 .

We're asked to approximate $\int_{0.05}^{0.35}14^{2\sqrt{x}}dx$

So the interval is [0.05,0.35] , the subintervals have length $\frac{0.35-0.05}{3}=0.1$ , and since we are using the midpoints of each interval, the x-values are [0.1,0.2,0.3]

$\int_{0.05}^{0.35}14^{2\sqrt{x}}dx\approx (0.1)[14^{2\sqrt{0.1}}+14^{2\sqrt{0.2}}+14^{2\sqrt{0.3}}]$

$\int_{0.05}^{0.35}14^{2\sqrt{x}}dx\approx 3.39$

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Example Question #179 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{-1.5}^{2.5}sin^3(4x^2)dx$ using four midpoints.

Possible Answers:

-2.03

-1.99

3.11

-0.89

1.88

Correct answer:

-0.89

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 asked to approximate $\int_{-1.5}^{2.5}sin^3(4x^2)dx$

So the interval is [-1.5,2.5] , the subintervals have length $\frac{2.5-(-1.5)}{4}=1$ , and since we are using the midpoints of each interval, the x-values are [-1,0,2,3]

$\int_{-1.5}^{2.5}sin^3(4x^2)dx\approx (1)[sin^3(4(-1)^2)+sin^3(4(0)^2)+sin^3(4(1)^2)+sin^3(4(2)^2)]$

$\int_{-1.5}^{2.5}sin^3(4x^2)dx\approx -0.89$

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Example Question #180 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{1.5}^{4.5}sin^2(\sqrt{4x})dx$ using three midpoints.

Possible Answers:

1.59

-1.29

0.23

-0.48

1.02

Correct answer:

1.02

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 asked to approximate $\int_{1.5}^{4.5}sin^2(\sqrt{4x})dx$

So the interval is [1.5,4.5] , the subintervals have length $\frac{4.5-1.5}{3}=1$ , and since we are using the midpoints of each interval, the x-values are [2,3,4]

$\int_{1.5}^{4.5}sin^2(\sqrt{4x})dx\approx (1)[sin^2(\sqrt{4(1)})+sin^2(\sqrt{4(2)})+sin^2(\sqrt{4(3)})]$

$\int_{1.5}^{4.5}sin^2(\sqrt{4x})dx\approx 1.02$

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Example Question #181 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{-1.5}^{1.5}6e^{x\sqrt{-\pi^2 x^2}}dx$ using three midpoints.

Possible Answers:

$3\pi-i$

$-3\pi+i$

Correct answer:

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 asked to approximate $\int_{-1.5}^{1.5}6e^{x\sqrt{-\pi^2 x^2}}dx$

So the interval is [-1.5,1.5] , the subintervals have length $\frac{1.5-(-1.5)}{3}=1$ , and since we are using the midpoints of each interval, the x-values are [-1,0,1]

$\int_{-1.5}^{1.5}6e^{x\sqrt{-\pi^2 x^2}}dx\approx (1)[6e^{(-1)\sqrt{-\pi^2 (-1)^2}}+6e^{(0)\sqrt{-\pi^2 (0)^2}}+6e^{(1)\sqrt{-\pi^2 (1)^2}}]$

$\int_{-1.5}^{1.5}6e^{x\sqrt{-\pi^2 x^2}}dx\approx -6$

Now, the complex number in the exponent may've been a point of curiosity before a calculator was used to find the result. However, this is a good opportunity to point out a unique mathematical identity: Euler's Identity. Euler's Identity, or rather a facet of it states:

$cos(x)=\frac{e^{-ix}+e^{ix}}{2}$

$2cos(x)=e^{-ix}+e^{ix}$

In the case of our problem sum, two of our three elements represent a function of this form:

$6e^{-i\pi}+6e^{i\pi}=2\cdot6cos(\pi)=-12$

Mathematical beauty.

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #172 : Midpoint Riemann Sums

Example Question #173 : Midpoint Riemann Sums

Example Question #174 : Midpoint Riemann Sums

Example Question #175 : Midpoint Riemann Sums

Example Question #176 : Midpoint Riemann Sums

Example Question #177 : Midpoint Riemann Sums

Example Question #178 : Midpoint Riemann Sums

Example Question #179 : Midpoint Riemann Sums

Example Question #180 : Midpoint Riemann Sums

Example Question #181 : Midpoint Riemann Sums

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