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Calculus 1 : How to find midpoint Riemann sums

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Example Questions

Example Question #31 : Midpoint Riemann Sums

Approximate the integral of the function $f(x)=2+2e^{(x-1)^2}$ for x =1 to using midpoint Reimann sums and two midpoints.

Possible Answers:

2085.19

32431.21

154724.12

230.39

485.22

Correct answer:

485.22

Explanation:

The form of a Riemann sum follows:

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

Since two mid points are asked for, divide the interval into two: [1,2.5][2.5,4] , and find the mid points of those: [1.75,3.25] .

Therefore, the approximation of the integral is:

$\frac{4-1}{2}((2+2e^{(1.75-1)^2})+(2+2e^{(3.25-1)^2}))$

$\frac{3}{2}(4+2e^{(0.75)^2}+2e^{(2.25)^2})$

485.22

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Example Question #32 : How To Find Midpoint Riemann Sums

Approximate the integral $\int_{0}^{4}cos(e^x)dx$ using the method of midpoint Reimann sums and two midpoints.

Possible Answers:

0.484

-0.332

0.242

-1.166

-0.166

Correct answer:

-1.166

Explanation:

The form of a Riemann sum follows:

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

For two intervals of length $\frac{4-0}{2}=2$ , the midpoints are [1,3] .

Therefore, the approximation of the integral is:

$\frac{4-0}{2}(cos(e^1)cos(e^3))$

-1.16627...

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Example Question #33 : How To Find Midpoint Riemann Sums

Approximate the integral of f(x)=x^2 for x = 2 to using midpoint Reimann sums and three midpoints.

Possible Answers:

232

112

166

Correct answer:

166

Explanation:

The form of a Riemann sum follows:

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

The interval [2,8] can be divided into three intervals of length $\frac{8-2}{3}=2$

[2,4][4,6][6,8] , with midpoints of [3,5,7] .

Therefore, the approximation of the integral is:

$\frac{8-2}{3}(3^2+5^2+7^2)$

166

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Example Question #34 : How To Find Midpoint Riemann Sums

Using midpoint Riemann sums and three mid points, approximate the area between the functions h(x)=x and $g(x)=\frac{x^2}{9}$ .

Possible Answers:

15.75

14.25

Correct answer:

14.25

Explanation:

The form of a Riemann sum follows:

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

It can be applied to combinations of functions, i.e. f(x)=h(x)-g(x)

To find the interval for this problem, find the intersections of the two functions, i.e. the values of that satisfy

$x=\frac{x^2}{9}$

$x\left(\frac{x}{9}-1\right)=0$

x=0,9

Knowing this interval, it can be divided into three intervals of length $\frac{9-0}{3}=3$ :

[0,3][3,6][6,9] , with midpoints [1.5,4.5,7.5] .

So the approximation of the area between functions is:

$\frac{9-0}{3}\left(\left(1.5-\frac{1.5^2}{9}\right)+\left(4.5-\frac{4.5^2}{9}\right)+\left(7.5-\frac{7.5^2}{9}\right)\right)$

14.25

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Example Question #31 : How To Find Midpoint Riemann Sums

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

Possible Answers:

2.090

3.337

3.704

0.834

0.926

Correct answer:

0.926

Explanation:

The form of a Riemann sum follows:

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

The interval $[0,\sqrt{\pi}]$ can be divided into four intervals of length

$\frac{\sqrt{\pi}-0}{4}=\frac{\sqrt{\pi}}{4}$ .

$\left[0,\frac{\sqrt{\pi}}{4}\right]\left[\frac{\sqrt{\pi}}{4},\frac{\sqrt{\pi}}{2}\right]\left[\frac{\sqrt{\pi}}{2},\frac{3\sqrt{\pi}}{4}\right]\left[\frac{3\sqrt{\pi}}{4},\pi\right]$

With the midpoints

$\left[\frac{\sqrt{\pi}}{8},\frac{3\sqrt{\pi}}{8},\frac{5\sqrt{\pi}}{8},\frac{7\sqrt{\pi}}{8}\right]$ .

Therefore, the approximation of the integral is:

$\frac{\sqrt{\pi}-0}{4}\left(sin\left(\left(\frac{\sqrt{\pi}}{8}\right)^2\right)+sin\left(\left(\frac{3\sqrt{\pi}}{8}\right)^2\right)+sin\left(\left(\frac{5\sqrt{\pi}}{8}\right)^2\right)+sin\left(\left(\frac{7\sqrt{\pi}}{8}\right)^2\right)\right)$

0.926

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Example Question #1061 : Calculus

Find the midpoint Reimann sum for the integral of the function $f(x)=e^{cos(x)}$ over the interval $x=[0,2\pi]$ using four midpoints.

Possible Answers:

$\frac{\pi}{2}(e^{\frac{1}{\sqrt2}}+e^{-\frac{1}{\sqrt2}})$

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

$\pi(2e^{\frac{1}{\sqrt2}}+2e^{-\frac{1}{\sqrt2}})$

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

$\pi(e^{\frac{1}{\sqrt2}}+e^{-\frac{1}{\sqrt2}})$

Correct answer:

$\pi(e^{\frac{1}{\sqrt2}}+e^{-\frac{1}{\sqrt2}})$

Explanation:

The form of a Riemann sum follows:

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

The interval $[0,2\pi]$ can be divided into four intervals of length $\frac{2\pi-0}{4}=\frac{\pi}{2}$

with mid points $\left[\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\right]$ .

The Riemann sum is then

$\frac{2\pi-0}{4}(e^{cos(\frac{\pi}{4})}+e^{cos(\frac{3\pi}{4})}+e^{cos(\frac{5\pi}{4})}+e^{cos(\frac{7\pi}{4})})$

$\pi(e^{\frac{1}{\sqrt2}}+e^{-\frac{1}{\sqrt2}})$

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Example Question #34 : How To Find Midpoint Riemann Sums

Use midpoint Riemann Sums to approximate the area between the -axis and the function $f(x)=\frac{1}{4}x^2 + 2$ for $-2 \leq x \leq 5$ . Use seven intervals n=7 .

Possible Answers:

24.9375

10.6875

10.9375

22.75

22.875

Correct answer:

24.9375

Explanation:

First, graph the equation to see if it is above the x-axis - positive, or below - negative. Then we can visualize the intervals and their midpoints:

Riemann 1

Dividing the region from x=-2 to x=5 into 7 intervals gives each a length $\Delta x$ of exactly 1.

We will be calculating the sum $A \approx \sum_{7}^{n=1}f(x_{n})*\Delta x$ where $x{_{n}}$ is the nth interval's midpoint. Since $\Delta x$ is jus 1, in this case we're just adding together this sum:

f(-1.5)+f(-0.5)+f(0.5)+f(1.5)+f(2.5)+f(3.5)+f(4.5)

Evaluating at these points and adding the results together yields 24.9375

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Example Question #32 : Midpoint Riemann Sums

Use Riemann midpoint sums to approximate the area between the curve $g(x)=\frac{1}{10}x^3 - \frac{1}{2}x+4$ and the -axis between x=0 and x=3 . Use five intervals n=5 .

Possible Answers:

15.8825

9.5295

18.9013

11.7345

19.5575

Correct answer:

11.7345

Explanation:

We are looking to approximate the area between 0 and 3, and we are dividing it into 5 intervals, so each interval will have a length of $\Delta x=\frac{3-0}{5}=0.6$ .

The midpoints of the 5 intervals are at x=0.3, 0.9, 1.5, 2.1, 2.7 , so we'll be evaluating the function for these values of x.

Riemann 2

The formula for the Riemann midpoint sum is $A \approx \sum_{n=1}^{5}f(x_{n})*\Delta x$ .

Note that you get the same answer regardless if you multiply each individual $f(x_{n})$ times $\Delta x$ or calculate the sum first then multiply by 0.6.

In this case, adding together the heights at the midpoints and multiplying by 0.6 yields 11.7345 .

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Example Question #1062 : Calculus

Consider the function y=ln(sin(x)) between x=1 and x= 2 . Find the definite integral using midpoint Riemann sums with two rectangles.

Possible Answers:

$\frac{[ln(sin(1.25)+ln(sin(1.75))]}{2}$

$\frac{ln(sin(3))}{2}$

$\frac{ln(3)}{2}$

Correct answer:

$\frac{[ln(sin(1.25)+ln(sin(1.75))]}{2}$

Explanation:

Midpoint sums are given by the formula

$\int_{a}^{b}f(x)\approx \frac{b-a}{n}*[f(x_{1})+f(x_2)]$ ,

where n is the number of steps and the x values are the midpoints of the rectangles.

Since we have 2 steps, there would be 2 rectangles, from x: (1,1.5) and the other one x:(1.5, 2) . The heights of these rectangles are calculated at the midpoint of the two ends, which occur at x=1.25 and x=1.75

In our case, the rectangle midpoints are at 1.25 and 1.75, since we only have 2 steps.

Therefore, the value is

$\frac{2-1}{2}*[ln(sin(1.25)+ln(sin(1.75))]=\frac{[ln(sin(1.25)+ln(sin(1.75))]}{2}$

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Example Question #40 : How To Find Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate the integral of the function f(x)=cos(x^2+x+1) over the interval $x=0,2\pi$ using four midpoints.

Possible Answers:

$\\ \left(cos\left(\frac{\pi^2}{16}+\left(\frac{\pi}{4}\right)+1\right)+cos\left(\frac{9\pi^2}{16}\right)+\left(\frac{3\pi}{4}\right)+1\right)+cos\left(\left(\frac{25\pi^2}{16}\right)+\left(\frac{5\pi}{4}\right)+1\right)$

$\\ \left(cos\left(\frac{\pi^2}{8}+\left(\frac{\pi}{2}\right)+1\right)+cos\left(\frac{9\pi^2}{16}\right)+\left(\frac{3\pi}{4}\right)+1\right)+cos\left(\left(\frac{25\pi^2}{16}\right)+\left(\frac{5\pi}{4}\right)+1\right)+cos\left(\left(\frac{49\pi^2}{16}\right)+\left(\frac{7\pi}{4}\right)+1\right)\right)$

$\\ \frac{\pi}{2}\left(cos\left(\frac{\pi^2}{4}+\left(\frac{\pi}{3}\right)+1\right)+cos\left(\frac{9\pi^2}{4}\right)+\left(\frac{3\pi}{2}\right)+1\right)+cos\left(\left(\frac{5\pi^2}{16}\right)+\left(\frac{5\pi}{4}\right)+1\right)+cos\left(\left(\frac{16\pi^2}{49}\right)+\left(\frac{7\pi}{4}\right)+1\right)\right)$

$\\ \frac{\pi}{2}\left(cos\left(\frac{\pi^2}{16}+\left(\frac{\pi}{4}\right)+1\right)+cos\left(\frac{9\pi^2}{16}\right)+\left(\frac{3\pi}{4}\right)+1\right)+cos\left(\left(\frac{25\pi^2}{16}\right)+\left(\frac{5\pi}{4}\right)+1\right)+cos\left(\left(\frac{49\pi^2}{16}\right)+\left(\frac{7\pi}{4}\right)+1\right)\right)$

Correct answer:

Explanation:

The general formula for a Reimann with n intervals from a to b is:

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

The term $\frac{b-a}{n}$ is also the length of an interval, so for this problem, an interval has a length of:

$\frac{2\pi-0}{4}=\frac{\pi}{2}$

And the four midpoints are thus:

$\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$

And the midpoint Reimann sum is:

$\frac{\pi}{2}\left(cos\left(\left(\frac{\pi}{4}\right)^2+\left(\frac{\pi}{4}\right)+1\right)+cos\left(\left(\frac{3\pi}{4}\right)^2+\left(\frac{3\pi}{4}\right)+1\right)+cos\left(\left(\frac{5\pi}{4}\right)^2+\left(\frac{5\pi}{4}\right)+1\right)+cos\left(\left(\frac{7\pi}{4}\right)^2+\left(\frac{7\pi}{4}\right)+1\right)\right)$

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Calculus 1 : How to find midpoint Riemann sums

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #31 : Midpoint Riemann Sums

Example Question #32 : How To Find Midpoint Riemann Sums

Example Question #33 : How To Find Midpoint Riemann Sums

Example Question #34 : How To Find Midpoint Riemann Sums

Example Question #31 : How To Find Midpoint Riemann Sums

Example Question #1061 : Calculus

Example Question #34 : How To Find Midpoint Riemann Sums

Example Question #32 : Midpoint Riemann Sums

Example Question #1062 : Calculus

Example Question #40 : How To Find Midpoint Riemann Sums

All Calculus 1 Resources

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