Midpoint Riemann Sums - AP Calculus AB

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Question

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

Answer

The Reimann sum approximation of an integral of a function with subintervals over an interval 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 .

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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Question

A Riemann Sum approximation of an integral $\int_a^b (f(x))dx$ follows the form

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

Where n is number of points/subintervals used to approximate the integral.

Knowing this, imagine a modified style of Riemann Sum, such that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

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:

$\int_{10}^{100} \frac{e^x}{x^2}dx$ ?

Answer

The more points/intervals that are selected, the closer the Reimann sum approximation becomes to the actual integral value, so n=20 will give a closer approximation than n=10:

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.

Consider the function $\frac{e^x}{x^2}$ over the interval . The steepness of this graph at a point can be found by taking the function's derivative.

The quotient rule of derivatives states:

$d\frac{u}{v}=\frac{vdu-udv}{v^2}$

So the derivative is

$\frac{x^2e^x-2xe^x}{x^4}$

$\frac{e^x(x-2)}{x^3}$

It might be intuitive to see that the steepness is positive and that it gets progressively greater over the specified interval, although to be precise, we may take the derivative once more to find the rate of change of this steepness and see if it's postive or negative:

$\frac{x^4(2xe^x+x^2e^x-2e^x-2xe^x)-4x^3(x^2e^x-2xe^x)}{x^8}$

$\frac{x^6e^x+6x^4e^x-4x^5e^x}{x^8}$

$\frac{e^x(x^2-4x+6)}{x^4}$

Since this derivative is positive over the specified interval, , and since the slope was shown to be positive as well, the slope is ever-increasing and growing steeper.

The intervals should in turn grow increasingly thinner.

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Question

The general Reimann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

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

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

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:

$\int_0^2 (x^3-6x^2+12x+6)dx$ ?

Answer

The more points/intervals that are selected, the closer the Reimann sum approximation becomes to the actual integral value, so n=50 will give a closer approximation than n=40:

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.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

First find the derivative of the function :

This can be rewritten as

Thus for all values for x outside of 2, the slope is positive, though as the function is continuous, a single point is nigh negligible.

Now take the second derivative:

For the interval , the second derivative is negative.

Since the signs of the first and second derivatives are opposing, the function is flattening out. The intervals should grow wider.

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Question

The general Reimann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

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

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

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:

$\int_{0.1}^{0.01} x\ln(x)dx$ ?

Answer

The more points/intervals that are selected, the closer the Reimann sum approximation becomes to the actual integral value, so will give a closer approximation than :

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.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

For the function

$x\ln(x)$

The derivative can be found, using the product rule :

$\ln(x)+x\frac{1}{x}$

$\ln(x)+1$

Over the entire interval , this function is negative. Now, consider the second derivative:

$\frac{1}{x}$

Over the interval , the function is positive.

Since the first and second derivatives have opposing signs, the function is flattening out. The subintervals should grow progressively thicker.

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Question

The general Riemann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

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

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

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:

$\int_0^{0.5} \frac{1}{2}^{x^2-2x+1}dx$ ?

Answer

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:

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.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

Considering $f(x)=\frac{1}{2}^{x^2-2x+1}$ , find the first derivative:

$f'(x)=(2x-2)\frac{1}{2}^{x^2-2x+1}ln(\frac{1}{2})$

The term in the parenthesis is what may lend to variability in the sign. Over the interval the parenthetical is negative, so the derivative is positive, since $ln(\frac{1}{2})=-0.693$ , and the product of two negatives is positive. Now consider the second derivative:

$f''(x)=\frac{1}{2}^{x^2-2x+2}ln(\frac{1}{2})+(2x-2)^2\frac{1}{2}^{x^2-2x+1}ln^2(\frac{1}{2})$

The second term will always be positive, and the first term will always be negative. On the interval , it can be found, however, that the second derivative will be positive. If calculating this is to nettlesome, plot this function on the interval:

Since the first and second derivatives have opposite signs, the function is getting flatter. The intervals should in turn get wider.

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Question

The general Riemann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

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

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

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:

$\int_0^{\frac{\pi}{2}} cos(\frac{x^2}{\pi})dx$ ?

Answer

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:

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.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

Considering $f(x)=cos(\frac{x^2}{\pi})$ take the first derivative:

$f'(x)=-\frac{2x}{\pi}sin(\frac{x^2}{\pi})$

Within the interval $x=[0,\frac{\pi}{2}]$ this function is negative. Now consider the second derivative:

$f''(x)=-\frac{2}{\pi}sin(\frac{x^2}{\pi})-\frac{4x^2}{\pi^2}cos(\frac{x^2}{\pi})$

Over the interval $x=[0,\frac{\pi}{2}]$ , the second derivative is negative as well.

Since the signs match, the slope gets steeper, and the intervals should get thinner.

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Question

The general Riemann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

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

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

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:

$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} e^{cos(x)}dx$ ?

Answer

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:

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.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

Considering $f(x)= e^{cos(x)}$ , find the first derivative:

$f'(x)=-sin(x)e^{cos(x)}$

Within the interval $x=[\frac{\pi}{3},\frac{\pi}{2}]$ , the function only takes negative values. Now, take the second derivative:

$f''(x)=-cos(x)e^{cos(x)}+sin^2(x)e^{cos(x)}$

On the interval $x=[\frac{\pi}{3},\frac{\pi}{2}]$ , this function takes on positive values.

Since the signs of the derivatives are opposing, the function gets flatter over the course of this interval, and the subintervals should thus get wider.

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Question

The general Riemann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

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

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

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:

$\int_{ln(\frac{\pi}{2})}^{ln(\pi)} sin(e^{x})dx$ ?

Answer

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:

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.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

Considering the function $f(x)=sin(e^{x})$ , take the first derivative:

$f'(x)=e^xcos(e^{x})$

Over the interval $x=[ln(\frac{\pi}{2}),ln(\pi)]$ this function is negative due to the cosine term. Now take the second derivative:

$f''(x)=e^xcos(e^{x})-e^{2x}sin(x)$

This function is also negative over the interval $x=[ln(\frac{\pi}{2}),ln(\pi)]$ .

Since both derivatives share a sign, the function grows steeper over the specified interval. The subintervals should therefore get progressively narrower.

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Question

Let

What is the Midpoint Riemann Sum on the interval divided into four sub-intervals?

Answer

The interval divided into four sub-intervals gives rectangles with vertices of the bases at

For the Midpoint Riemann sum, we need to find the rectangle heights which values come from the midpoint of the sub-intervals, or f(1), f(3), f(5), and f(7).

Because each interval has width 2, the approximated Midpoint Riemann Sum is

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Question

Estimate the area under the curve for the following function using a midpoint Riemann sum from to with .

$f(x)=\frac{1}{2}x^3+4$

Answer

If we want to estimate the area under the curve from to and are told to use , this means we estimate the area using two rectangles that will each be two units wide and whose height is the value of the function at the midpoint of the interval. We have a rectangle from to , whose height is the value of the function at , and a rectangle from to , whose height is the value of the function at . First we can find the value of the function at these midpoints, and then add the areas of the two rectangles, which gives us the following:

$f(1)=\frac{1}{2}(1)^3+4=\frac{9}{2}$

$f(3)=\frac{1}{2}(3)^3+4=\frac{35}{2}$

$A=2\left(\frac{9}{2}\right)+2\left(\frac{35}{2}\right)=44$

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Question

Estimate the area under the curve for the following function from to using a midpoint Riemann sum with rectangles:

$f(x)=\frac{1}{4}x^2+3$

Answer

If we are told to use rectangles from to , this means we have a rectangle from to , a rectangle from to , a rectangle from to , and a rectangle from to . We can see that the width of each rectangle is because we have an interval that is units long for which we are using rectangles to estimate the area under the curve. The height of each rectangle is the value of the function at the midpoint for its interval, so first we find the height of each rectangle and then add together their areas to find our answer:

$f(0.5)=\frac{1}{4}(0.5)^2+3=\frac{49}{16}$

$f(1.5)=\frac{1}{4}(1.5)^2+3=\frac{57}{16}$

$f(2.5)=\frac{1}{4}(2.5)^2+3=\frac{73}{16}$

$f(3.5)=\frac{1}{4}(3.5)^2+3=\frac{97}{16}$

$A=(1)\left(\frac{49}{16}\right)+(1)\left(\frac{57}{16}\right)+(1)\left(\frac{73}{16}\right)+(1)\left(\frac{97}{16}\right)=\frac{276}{16}=\frac{69}{4}$

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Question

Using 4 intervals, calculate the midpoint Riemann sum approximation of the area under the curve defined by $\small f(x) = 3x^2 + 5x + 6$ on the interval $\small [0, 8]$ .

Answer

To make 4 intervals, we need 5 boundary lines. In this case, they are the vertical lines at $\small x = 0, 2, 4, 6, 8$ . So the midpoints of each interval are at $\small x = 1, 3, 5, 7$ , respectively.

The area of each region is found by $\small f(x)\cdot \Delta x$ , where $\small \Delta x$ is the width of the interval (here, 2). So we simply need to find the functional value at each $\small x$ -coordinate of the midpoints above, multiply by 2, and sum. Equivalently, we could sum the functional values, then double the sum.

$\small f(1) = 14$

$\small f(3) = 48$

$\small f(5) = 106$

$\small f(7) = 188$

Their sum is $\small 356$ , and twice that is $\small 712$ .

The wrong answer $\small 504$ comes from using $\small x = 0, 2, 4, 6$ as the midpoints.

The wrong answer $\small 968$ comes from using $\small x = 2, 4, 6, 8$ as the midpoints.

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Question

Find the area under on the interval using five midpoint Riemann sums.

Answer

The problem becomes this:

Addings these rectangles up to approximate the area under the curve is

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Question

Approximate the area under the curve from $0\leq x\leq 8$ using the midpoint Riemann Sum with a partition of size five given the graph of the function.

Answer

We begin by finding the given change in x:

$\Delta x=\frac{b-a}{n}=\frac{8}{5}=1.6$

We then define our partition intervals:

$[x_{i-1},x_i]=[0,1.6),[1.6,3.2),[3.2,4.8),[4.8,6.4),[6.4,8]$

We then choose the midpoint in each interval:

${\bar{x_i}}={0.8,2.4,4.0,5.6,7.2}$

Then we find the value of the function at the point. This is determined through observation of the graph

$f(\bar{x_i})\approx 4.8,4.5,4,1.6,3.4$

Then we simply substitute these values into the formula for the Riemann Sum

$\A\approx \sum_{i=1}^nf(x_i^*)\Delta x=4.8\cdot 1.6+4.5\cdot 1.6+4\cdot 1.6+1.6\cdot 1.6+3.4\cdot 1.6\ =1.6(4.8+4.5+4+1.6+3.4)=1.6\cdot 18.3=29.28$

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Question

Approximate the area underneath the given curve using the Riemann Sum with eight intervals for $-4\leq x\leq12$ .

Answer

We begin by defining the size of our partitions and the partitions themselves.

$\ \Delta x=\frac{b-a}{n}=\frac{12+4}{8}=2\$

$[x_{i-1},x_i]=[-4,-2],[-2,0],[0,2],[2,4],[4,6],[6,8],[8,10],[10,12]$

We then choose the midpoint in each interval:

$\bar{x_i}=-3,-1,1,3,5,7,9,11$

Then we find the function value at each point.

$f(\bar{x_i})=-131,-21,1,31,165,499,1129,2151$

We then substitute these values into the Riemann Sum formula.

$A\approx\sum_{i=1}^nf(\bar{x_i})\Delta x=2(-131-21+1+31+165+499+1129+2151)=7648$

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Question

Solve the integral

$\int_{3}^{7}\frac{1}{x}dx$

using the midpoint Riemann sum approximation with subintervals.

Answer

Midpoint Riemann sum approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx (\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ]$

where is the number of subintervals and $f(m_{i})$ is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{n})=(\frac{7-3}{10})=0.4$ .

The approximate value at each midpoint is below.

The sum of all the approximate midpoints values is , therefore

$(\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ] =0.4(2.116738)\approx0.8467$

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Question

Solve the integral

$\int_{1}^{8}{\sqrt{x}}dx$

using the midpoint Riemann sum approximation with subintervals.

Answer

Midpoint Riemann sum approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx (\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ]$

where is the number of subintervals and $f(m_{i})$ is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{n})=(\frac{8-1}{10})=0.7$ .

The approximate value at each midpoint is below.

The sum of all the approximate midpoints values is , therefore

$(\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ] =0.7(20.6068259)\approx14.4248$

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Question

Solve the integral

$\int_{0}^{\pi /2}\cos (x)dx$

using the midpoint Riemann sum approximation with subintervals.

Answer

Midpoint Riemann sum approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx (\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ]$

where is the number of subintervals and $f(m_{i})$ is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{n})=(\frac{\pi /2-0}{10})=\pi/20\approx0.1571$ .

The approximate value at each midpoint is below.

The sum of all the approximate midpoints values is , therefore

$(\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ] =0.1571(6.37275)\approx1$

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Question

Solve the integral

$\int_{2}^{5}{e{^{x}}}dx$

using the midpoint Riemann sum approximation with subintervals.

Answer

Midpoint Riemann sum approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx (\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ]$

where is the number of subintervals and $f(m_{i})$ is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{n})=(\frac{5-2}{10})=0.3$ .

The approximate value at each midpoint is below.

The sum of all the approximate midpoints values is , therefore

$(\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ] =0.3(468.3221)\approx140.4966$

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Question

Solve the integral

$\int_{2}^{7}{\frac{1}{5x+3}}dx$

using the trapezoidal approximation with subintervals.

Answer

Trapezoidal approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx T_{n}=(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]$

where is the number of subintervals and is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{7-2}{25})=0.5$ .

The value of each approximation term is below.

The sum of all the approximation terms is , therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.5)(0.86027754)=0.4301$

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