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Calculus 1 : Functions

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Example Question #41 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral of the function $f(x)=x^{2x}$ over the interval x=[0.5,3.5] using three midpoints.

Possible Answers:

6535.0

746

248.7

812

101.5

Correct answer:

746

Explanation:

A Reimann sum integral approximation of a function f(x) , with points on the interval [a,b] follows the form:

$\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

For this problem, there will be three intervals of length $\frac{3.5-.5}{3}=1$ with midpoints [1,2,3] .

Therefore, the integral approximation of $\int_{.5}^{3.5}x^{2x}dx$ is:

(1^2+2^4+3^6)=1+16+729=746

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Example Question #42 : Midpoint Riemann Sums

Approximate the value of the integral of the function $f(x)=e^{cos(x)}$ on the interval $x=[0,2\pi]$ using the method of Midpoint Reimann Sums and four midpoint values.

Possible Answers:

$2e^{\sqrt2/2}+2 e^{-\sqrt2/2}$

$\frac{\pi}{2}(e+\frac{1}{e})$

$e+\frac{1}{e}$

$\pi e^{\sqrt2/2}+\pi e^{-\sqrt2/2}$

Correct answer:

$\pi e^{\sqrt2/2}+\pi e^{-\sqrt2/2}$

Explanation:

A Reimann sum integral approximation of a function f(x) , with points on the interval [a,b] follows the form:

$\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

For this problem, four midpoints means four intervals of length $\frac{2\pi-0}{4}=\frac{\pi}{2}$ with midpoints $[\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}]$

The approximation for this problem is thus:

$\frac{\pi}{2}(e^{cos(\pi/4)}+e^{cos(3\pi/4)}+e^{cos(5\pi/4)}+e^{cos(7\pi/4)})$

$\frac{\pi}{2}(e^{\sqrt2/2}+e^{-\sqrt2/2}+e^{-\sqrt2/2}+e^{\sqrt2/2})$

$\pi e^{\sqrt2/2}+\pi e^{-\sqrt2/2}$

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Example Question #43 : Midpoint Riemann Sums

Using the method of Midpoint Reimann sums, approximate the integral of f(x)=x^2+x-6 over the inteval x=[6,10] using two midpoints.

Possible Answers:

308

134

268

102

204

Correct answer:

268

Explanation:

A Reimann sum integral approximation, for points/intervals over the range [a,b] , follows the form:

$\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

In the case of the problem, the intervals will have length $\frac{10-6}{2}=2$ , with midpoints [7,9] .

For the function f(x)=x^2+x-6 , the Reimann sum approximation is then:

2((49+7-6)+(81+9-6))

2(50+84)

268

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Example Question #44 : Midpoint Riemann Sums

This table gives values of a function f(x) at certain values of .

Riemann sum problem 2

Using the table above, find the midpoint Riemann sum of f(x) with n=3 from x=0 to x=6 .

Possible Answers:

24.5

Correct answer:

Explanation:

$\Delta x = \frac {6-0}{3} = 2$

Thus, our intervals are to , to , and to .

The midpoints of each interval are, respectively, , , and

Next, use the data table to find the values of f(x) at each value of :

f(1)=19

f(3)= 17

f(5)= 13

Finally, calculate the estimation of the area using these values and $\Delta x = 2$

2(19+17+13)

$\textbf{=98}$

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Example Question #45 : Midpoint Riemann Sums

This table gives values of a function f(x) at certain values of .

Midpoint riemann sum problem 3

Using the table above, find the midpoint Riemann sum of f(x) with n=2 from x=2 to x=10 .

Possible Answers:

120

Correct answer:

120

Explanation:

$\Delta x = \frac {10-2}{2} = 4$

Thus, our intervals are to and to .

The midpoints of each interval are, respectively, and

Next, use the data table to find the values of f(x) at each value of :

f(4)=6

f(8)=24

Finally, calculate the estimation of the area using these values and $\Delta x = 4$ :

4(6+24)

$\textbf{=120}$

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Example Question #46 : Midpoint Riemann Sums

Estimate the area under the curve for the function below with the midpoint Riemann sum, in the interval from x=1 to x=5 , using four rectangles. Round to the nearest hundredth.

$\frac{10}{3+x}$

Possible Answers:

6.91

6.35

7.61

7.11

Correct answer:

6.91

Explanation:

Since we're using the midpoint sum, we need to separate our interval into 4 subintervals for each rectangle: [1,2]; [2,3]; [3,4]; and [4,5]. Each of these is 1 unit across, and the height of the rectangle is going to be the value of the function at the midpoints of these intervals.

Since the area of a rectangle is length*height, we have the following sum as our approximation:

$\frac{10}{3+3/2}+\frac{10}{3+5/2}+\frac{10}{3+7/2}+\frac{10}{3+9/2}{}{}$

which comes out to

$\frac{8896}{1287}=6.9121989...{}$

We approximate this further as 6.91 and get our final answer.

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Example Question #47 : Midpoint Riemann Sums

Let f(x)=x .

What is the Midpoint Riemann sum on the interval [0,16] divided into four sub-intervals?

Possible Answers:

128

Correct answer:

128

Explanation:

The interval [0,16] divided into four sub-intervals gives rectangles with vertices of the bases at

x=0,4,8,12,16 .

For the Midpoint Riemann sum, we need to find the rectangle heights which values come from the midpoint of the sub-intervals, or f(2) , f(6) , f(10) , and f(14) .

f(2)=2

f(6)=6

f(10)=10

f(14)=14

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

A=bh=4(2+6+10+14)=128

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Example Question #48 : Midpoint Riemann Sums

Let f(x)=x^2+7x+10 .

What is the Midpoint Riemann sum on the interval [0,16] divided into four sub-intervals?

Possible Answers:

2400

1200

600

3000

Correct answer:

2400

Explanation:

the interval [0,16] divided into four sub-intervals gives rectangles with vertices of the bases at

x=0,4,8,12,16

For the Midpoint Riemann sum, we need to find the rectangle heights which values come from the midpoint of the sub-intervals, or f(2) , f(6) , f(10) , and f(14) .

f(2)=2^2+7*2+10=28

f(6)=6^2+7*6+10=88

f(10)=10^2+7*10+10=180

f(14)=14^2+7*14+10=304

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

A=bh=4(28+88+180+304)=2400

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Example Question #49 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral of the function $f(x)=sin(\frac{2}{\pi}x^2)$ over the interval $x=[0,4\pi]$ using four midpoints.

Possible Answers:

$2\pi$

$4\pi$

$\pi$

$16\pi$

Correct answer:

$4\pi$

Explanation:

The Reimann sum approximation of an integral of a function with subintervals over an interval [a,b] takes the form:

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

Where $\frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are four midpoints, the subintervals have length $\frac{4\pi-0}{4}=\pi$ , and the midpoints are $x=[\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2}]$ .

The integral is thus:

$\int_0^{4\pi} f(x)dx \approx \pi[sin(\frac{2}{\pi}(\frac{\pi}{2})^2)+sin(\frac{2}{\pi}(\frac{3\pi}{2})^2)+sin(\frac{2}{\pi}(\frac{5\pi}{2})^2)+sin(\frac{2}{\pi}(\frac{7\pi}{2})^2)]$

$\int_0^{4\pi} f(x)dx \approx \pi[sin(\frac{\pi}{2})+sin(\frac{9\pi}{2})+sin(\frac{25\pi}{2})+sin(\frac{49\pi}{2})]$

$\int_0^{4\pi} f(x)dx \approx \pi[1+1+1+1]$

$\int_0^{4\pi} f(x)dx \approx 4\pi$

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Example Question #50 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the area of the region between the functions $f(x)=10(1-e^{-x})$ and g(x)=2x over the interval x=[0,3] using three midpoints.

Possible Answers:

11.8825

12.4700

5.4413

8.9679

11.4979

Correct answer:

11.8825

Explanation:

The Reimann sum approximation of an integral of a function with subintervals over an interval [a,b] takes the form:

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

Where $\frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are three midpoints, the subintervals have length $\frac{3-0}{3}=1$ , and the midpoints are x=[0.5,1.5,2.5] .

Furthermore, since the question is asking for the area between the two functions, it's asking for the difference between the larger function over the interval, f(x), and the smaller function, g(x).

The integral is thus:

${\int_0^{3} (f(x)-g(x))dx \approx [(10(1-e^{-0.5})-2(0.5))+(10(1-e^{-1.5})-2(1.5))+(10(1-e^{-2.5})-2(2.5))]}$

${\int_0^{3} (f(x)-g(x))dx \approx 11.8825$

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #41 : Midpoint Riemann Sums

Example Question #42 : Midpoint Riemann Sums

Example Question #43 : Midpoint Riemann Sums

Example Question #44 : Midpoint Riemann Sums

Example Question #45 : Midpoint Riemann Sums

Example Question #46 : Midpoint Riemann Sums

Example Question #47 : Midpoint Riemann Sums

Example Question #48 : Midpoint Riemann Sums

Example Question #49 : Midpoint Riemann Sums

Example Question #50 : Midpoint Riemann Sums

All Calculus 1 Resources

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