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Calculus 1 : Differential Functions

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Example Question #91 : Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate the integral $\int_0^9 1.2^{x+1}$ using three midpoints.

Possible Answers:

27.04

22.16

15.13

31.90

9.01

Correct answer:

27.04

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 are approximating $\int_0^9 1.2^{x+1}$

So the interval is [0,9] , the subintervals have length $\frac{9-0}{3}=3$ , and since we are using the midpoints of each interval, the x-values are [1.5,4.5,7.5]

$\int_0^9 1.2^{x+1}\approx 3[1.2^{1.5+1}+1.2^{4.5+1}+1.2^{7.5+1}]$

$\int_0^9 1.2^{x+1}\approx 27.04$

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Example Question #92 : Midpoint Riemann Sums

Approximate the integral $\int_3^7 1.02^{x^3}$ using the method of midpoint Riemann sums and four midpoints.

Possible Answers:

23.7

8.9

39.2

11.2

265.4

Correct answer:

265.4

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 are approximating $\int_3^7 1.02^{x^3}$

So the interval is [3,7] , the subintervals have length $\frac{7-3}{4}=1$ , and since we are using the midpoints of each interval, the x-values are [3.5,4.5,5.5,6.5]

$\int_3^7 1.02^{x^3}\approx 1[1.02^{3.5^3}+1.02^{4.5^3}+1.02^{5.5^3}+1.02^{6.5^3}]$

$\int_3^7 1.02^{x^3} \approx 265.4$

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Example Question #93 : Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate the integral $\int_{-1.2}^{2.4}cos(1.7^{x^{1.3}})$ using four midpoints.

Possible Answers:

0.149+0.115i

0.149-0.115i

0.115-0.149i

0.115+0.149i

0.231+i

Correct answer:

0.115+0.149i

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 $\int_{-1.2}^{2.4}cos(1.7^{x^{1.3}})$

So the interval is [-1.2,2.4] , the subintervals have length $\frac{2.4-(-1.2)}{4}=0.9$ , and since we are using the midpoints of each interval, the x-values are [-0.75,0.15,1.05,1.95]

${\int_{-1.2}^{2.4}cos(1.7^{x^{1.3}})\approx 0.9[cos(1.7^{(-0.75)^{1.3}})+cos(1.7^{(0.15)^{1.3}})+cos(1.7^{(1.05)^{1.3}})+cos(1.7^{(1.95)^{1.3}})]}$

$\int_{-1.2}^{2.4}cos(1.7^{x^{1.3}}) \approx 0.115+0.149i$

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Example Question #94 : Midpoint Riemann Sums

Using the method of midpoint of Reimann sums, approximate the integral $\int_0^3 tan(x^2)dx$ using three midpoints.

Possible Answers:

0.49

2.31

-0.81

0.78

-1.02

Correct answer:

-1.02

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 $\int_0^3 tan(x^2)dx$

So the interval is [0,3] , the subintervals have length $\frac{3-0}{3}=1$ , and since we are using the midpoints of each interval, the x-values are [0.5,1.5,2.5]

$\int_0^3 tan(x^2)dx \approx 1[ tan(0.5^2)+ tan(1.5^2)+ tan(2.5^2)]=-1.02$

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Example Question #95 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_1^5 ln(tan(2^x))dx$ using four midpoints.

Possible Answers:

9.42-0.64 i

-0.64 -9.42i

-9.42-0.64 i

9.42+0.64 i

-0.64 +9.42i

Correct answer:

-0.64 +9.42i

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 $\int_1^5 ln(tan(2^x))dx$

So the interval is [1,5] , the subintervals have length $\frac{5-1}{4}=1$ , and since we are using the midpoints of each interval, the x-values are [1.5,2.5,3.5,4.5]

$\int_1^5 ln(tan(2^x))dx \approx 1[ln(tan(2^{1.5}))+ln(tan(2^{2.5}))+ln(tan(2^{3.5}))+ln(tan(2^{4.5}))]$

$\int_1^5 ln(tan(2^x))dx \approx -0.64 +9.42i$

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Example Question #91 : Functions

Using the method of midpoint Reimann sums, approximate the integral $\int_2^8 \frac{1}{tan(3^x)}dx$ using two midpoints.

Possible Answers:

6.1

-19.5

18.3

4.5

-6.5

Correct answer:

-19.5

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 $\int_2^8 \frac{1}{tan(3^x)}dx$

So the interval is [2,8] , the subintervals have length $\frac{8-2}{2}=3$ , and since we are using the midpoints of each interval, the x-values are [3.5,6.5]

$\int_2^8 \frac{1}{tan(3^x)}dx \approx 3[\frac{1}{tan(3^{3.5})}+\frac{1}{tan(3^{6.5})}]=-19.5$

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Example Question #91 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_3^9 \frac{3^{x}}{2^{x}}dx$ using three midpoints.

Possible Answers:

28.1

42.1

192.6

96.3

84.2

Correct answer:

84.2

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're approximating $\int_3^9 \frac{3^{x}}{2^{x}}dx$

So the interval is [3,9] , the subintervals have length $\frac{9-3}{3}=2$ , and since we are using the midpoints of each interval, the x-values are [4,6,8]

$\int_3^9 \frac{3^{x}}{2^{x}}dx \approx 2[\frac{3^{4}}{2^{4}}+\frac{3^{6}}{2^{6}}+\frac{3^{8}}{2^{8}}]=84.2$

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Example Question #91 : How To Find Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_1^{2.5} tan(\frac{1}{x^{1.4}})$ using three midpoints.

Possible Answers:

1.722

0.132

0.348

0.696

0.861

Correct answer:

0.861

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 $\int_1^{2.5} tan(\frac{1}{x^{1.4}})$

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]

$\int_1^{2.5} tan(\frac{1}{x^{1.4}})\approx 0.5[tan(\frac{1}{1.25^{1.4}})+tan(\frac{1}{1.75^{1.4}})+tan(\frac{1}{2.25^{1.4}})]$

$\int_1^{2.5} tan(\frac{1}{x^{1.4}})\approx 0.861$

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Example Question #99 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{2.5}^{4.5} \frac{cos(x^3)}{tan(x^4)}dx$ using four midpoints.

Possible Answers:

1.904

0.952

3.808

1.656

3.312

Correct answer:

1.904

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 $\int_{2.5}^{4.5} \frac{cos(x^3)}{tan(x^4)}dx$

So the interval is [2.5,4.5] , the subintervals have length $\frac{4.5-2.5}{4}=0.5$ , and since we are using the midpoints of each interval, the x-values are [2.75,3.25,3.75,4.25]

$\int_{2.5}^{4.5} \frac{cos(x^3)}{tan(x^4)}dx \approx 0.5[\frac{cos(2.75^3)}{tan(2.75^4)}+\frac{cos(3.25^3)}{tan(3.25^4)}+\frac{cos(3.75^3)}{tan(3.75^4)}+\frac{cos(4.25^3)}{tan(4.25^4)}]$

$\int_{2.5}^{4.5} \frac{cos(x^3)}{tan(x^4)}dx \approx 1.904$

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Example Question #100 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{10.5}^{13.5} tan(e^{x^{1.5}})dx$ using three midpoints.

Possible Answers:

1.114

0.229

-1.083

-0.014

0.782

Correct answer:

0.782

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 $\int_{10.5}^{13.5} tan(e^{x^{1.5}})dx$

So the interval is [10.5,13.5] , the subintervals have length $\frac{13.5-10.5}{3}=1$ , and since we are using the midpoints of each interval, the x-values are [11,12,13]

$\int_{10.5}^{13.5} tan(e^{x^{1.5}})dx \approx 1[tan(e^{11^{1.5}})+tan(e^{12^{1.5}})+tan(e^{13^{1.5}})]$

$\int_{10.5}^{13.5} tan(e^{x^{1.5}})dx \approx 0.782$

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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #91 : Midpoint Riemann Sums

Example Question #92 : Midpoint Riemann Sums

Example Question #93 : Midpoint Riemann Sums

Example Question #94 : Midpoint Riemann Sums

Example Question #95 : Midpoint Riemann Sums

Example Question #91 : Functions

Example Question #91 : Midpoint Riemann Sums

Example Question #91 : How To Find Midpoint Riemann Sums

Example Question #99 : Midpoint Riemann Sums

Example Question #100 : Midpoint Riemann Sums

All Calculus 1 Resources

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