Math Articles

Article / Updated 09-22-2022

When g'(x) = f(x), you can use the substitution u = g(x) to integrate expressions of the form f(x) multiplied by h(g(x)), provided that h is a function that you already know how to integrate. Variable substitution helps to fill the gaps left by the absence of a Product Rule and a Chain Rule for integration. Here’s a hairy-looking integral that actually responds well to substitution: The key insight here is that the numerator of this fraction is the derivative of the inner function in the denominator. Watch how this plays out in this substitution: Declare u equal to the inner function in the denominator and make the substitution: Here’s the substitution: Differential du = (2x + 1) dx: The second part of the substitution now becomes clear: Notice how this substitution hinges on the fact that the numerator is the derivative of the inner function in the denominator. (You may think that this is quite a coincidence, but coincidences like these happen all the time on exams!) Integration is now quite straightforward: You take an extra step to remove the fraction before you integrate: Substitute back x2 + x – 5 for u: Checking the answer by differentiating with the Chain Rule reveals how this problem was set up in the first place: Here’s another example where you make a variable substitution: Notice that the derivative of x4 – 1 is x3, off by a constant factor. So here’s the declaration, followed by the differentiation: Now you can just do both substitutions at once: At this point, you can solve the integral simply. Similarly, here’s another example: At first glance, this integral looks just plain horrible. But on further inspection, notice that the derivative of cot x is –csc2 x, so this looks like another good candidate: This results in the following substitution: Again, this is another integral that you can solve.

Article / Updated 09-22-2022

You can calculate a confidence interval (CI) for the mean, or average, of a population even if the standard deviation is unknown or the sample size is small. When a statistical characteristic that’s being measured (such as income, IQ, price, height, quantity, or weight) is numerical, most people want to estimate the mean (average) value for the population. You estimate the population mean, by using a sample mean, plus or minus a margin of error. The result is called a confidence interval for the population mean, In many situations, you don’t know so you estimate it with the sample standard deviation, s. But if the sample size is small (less than 30), and you can’t be sure your data came from a normal distribution. (In the latter case, the Central Limit Theorem can’t be used.) In either situation, you can’t use a z*-value from the standard normal (Z-) distribution as your critical value anymore; you have to use a larger critical value than that, because of not knowing what is and/or having less data. The formula for a confidence interval for one population mean in this case is is the critical t*-value from the t-distribution with n – 1 degrees of freedom (where n is the sample size). The t-table The t*-values for common confidence levels are found using the last row of the t-table above. The t-distribution has a shape similar to the Z-distribution except it’s flatter and more spread out. For small values of n and a specific confidence level, the critical values on the t-distribution are larger than on the Z-distribution, so when you use the critical values from the t-distribution, the margin of error for your confidence interval will be wider. As the values of n get larger, the t*-values are closer to z*-values. To calculate a CI for the population mean (average), under these conditions, do the following: Determine the confidence level and degrees of freedom and then find the appropriate t*-value. Refer to the preceding t-table. Find the sample mean and the sample standard deviation (s) for the sample. Multiply t* times s and divide that by the square root of n. This calculation gives you the margin of error. Take plus or minus the margin of error to obtain the CI. The lower end of the CI is minus the margin of error, whereas the upper end of the CI is plus the margin of error. Here's an example of how this works For example, suppose you work for the Department of Natural Resources and you want to estimate, with 95 percent confidence, the mean (average) length of all walleye fingerlings in a fish hatchery pond. You take a random sample of 10 fingerlings and determine that the average length is 7.5 inches and the sample standard deviation is 2.3 inches. Because you want a 95 percent confidence interval, you determine your t*-value as follows: The t*-value comes from a t-distribution with 10 – 1 = 9 degrees of freedom. This t*-value is found by looking at the t-table. Look in the last row where the confidence levels are located, and find the confidence level of 95 percent; this marks the column you need. Then find the row corresponding to df = 9. Intersect the row and column, and you find t* = 2.262. This is the t*-value for a 95 percent confidence interval for the mean with a sample size of 10. (Notice this is larger than the z*-value, which would be 1.96 for the same confidence interval.) You know that the average length is 7.5 inches, the sample standard deviation is 2.3 inches, and the sample size is 10. This means Multiply 2.262 times 2.3 divided by the square root of 10. The margin of error is, therefore, Your 95 percent confidence interval for the mean length of all walleye fingerlings in this fish hatchery pond is (The lower end of the interval is 7.5 – 1.645 = 5.86 inches; the upper end is 7.5 + 1.645 = 9.15 inches.) Notice this confidence interval is wider than it would be for a large sample size. In addition to having a larger critical value (t* versus z*), the smaller sample size increases the margin of error, because n is in its denominator. With a smaller sample size, you don’t have as much information to “guess” at the population mean. Hence keeping with 95 percent confidence, you need a wider interval than you would have needed with a larger sample size in order to be 95 percent confident that the population mean falls in your interval. Now, say it in a way others can understand After you calculate a confidence interval, make sure you always interpret it in words a non-statistician would understand. That is, talk about the results in terms of what the person in the problem is trying to find out — statisticians call this interpreting the results “in the context of the problem.” In this example you can say: “With 95 percent confidence, the average length of walleye fingerlings in this entire fish hatchery pond is between 5.86 and 9.15 inches, based on my sample data.” (Always be sure to include appropriate units.)

Article / Updated 09-22-2022

You can use the sum and difference formulas for cosine to calculate the cosine of the sums and differences of angles similarly to the way you can use the sum and difference formulas for sine, because the formulas look very similar to each other. When working with sines and cosines of sums and differences of angles, you're simply plugging in given values for the variables (angles). Just make sure you use the correct formula based on the information you're given in the question. Here are the sum and difference formulas for cosines: The sum and difference formulas for cosine (and sine) can do more than calculate a trig value for an angle not marked on the unit circle (at least for angles that are multiples of 15 degrees). They can also be used to find the cosine (and sine) of the sum or difference of two angles based on information given about the two angles. For such problems, you'll be given two angles (call them A and B), the sine or cosine of A and B, and the quadrant(s) in which the two angles are located. Use the following steps to find the exact value of cos(A + B), given that cos A = –3/5, with A in quadrant II of the coordinate plane, and sin B = –7/25, with B in quadrant III: Choose the appropriate formula and substitute the information you know to determine the missing information. then substitutions result in this equation: To proceed any further, you need to find cos B and sin A. Draw pictures representing right triangles in the quadrant(s). Drawing pictures helps you visualize the missing pieces of info. You need to draw one triangle for angle A in quadrant II and one for angle B in quadrant III. Using the definition of sine as opp/hyp and cosine as adj/hyp, this figure shows these triangles. Notice that the value of a leg is missing in each triangle. To find the missing values, use the Pythagorean theorem. The length of the missing leg in Figure a is 4, and the length of the missing leg in Figure b is –24. Determine the missing trig ratios to use in the sum or difference formula. You use the definition of cosine to find that cos B = –24/25 and the definition of sine to find that sin A = 4/5. Substitute the missing trig ratios into the sum or difference formula and simplify. You now have this equation: Follow the order of operations to get this answer: This equation simplifies to cos(A + B) = 4/5.

Article / Updated 09-22-2022

You can find a confidence interval (CI) for the difference between the means, or averages, of two population samples, even if the population standard deviations are unknown and/or the sample sizes are small. The goal of many statistical surveys and studies is to compare two populations, such as men versus women, low versus high income families, and Republicans versus Democrats. When the characteristic being compared is numerical (for example, height, weight, or income), the object of interest is the amount of difference in the means (averages) for the two populations. For example, you may want to compare the difference in average age of Republicans versus Democrats, or the difference in average incomes of men versus women. You estimate the difference between two population means, by taking a sample from each population (say, sample 1 and sample 2) and using the difference of the two sample means plus or minus a margin of error. The result is a confidence interval for the difference of two population means, There are two situations where you cannot use z* when computing the confidence interval. The first of which is if you not know In this case you need to estimate them with the sample standard deviations, s1 and s2. The second situation is when the sample sizes are small (less than 30). In this case you can’t be sure whether your data came from a normal distribution. In either of these situations, a confidence interval for the difference in the two population means is where t* is the critical value from the t-distribution with n1 + n2 – 2 degrees of freedom; n1 and n2 are the two sample sizes, respectively; and s1 and s2 are the two sample standard deviations. This t*-value is found on the following t-table by intersecting the row for df = n1 + n2 – 2 with the column for the confidence level you need, as indicated by looking at the last row of the table. To calculate a CI for the difference between two population means, do the following: Determine the confidence level and degrees of freedom (n1 + n2 – 2) and find the appropriate t*-value. Refer to the above table. Identify Identify Find the difference, between the sample means. Calculate the confidence interval using the equation, Suppose you want to estimate with 95% confidence the difference between the mean (average) lengths of the cobs of two varieties of sweet corn (allowing them to grow the same number of days under the same conditions). Call the two varieties Corn-e-stats (group 1) and Stats-o-sweet (group 2). Assume that you don’t know the population standard deviations, so you use the sample standard deviations instead — suppose they turn out to be s1 = 0.40 and s2 = 0.50 inches, respectively. Suppose the sample sizes, n1 and n2, are each only 15. To calculate the CI, you first need to find the t*-value on the t-distribution with (15 + 15 – 2) = 28 degrees of freedom. Using the above t-table, you look at the row for 28 degrees of freedom and the column representing a confidence level of 95% (see the labels on the last row of the table); intersect them and you see t*28 = 2.048. For both groups, you took random sample of 15 cobs, with the Corn-e-stats variety averaging 8.5 inches, and Stats-o-sweet 7.5 inches. So the information you have is: The difference between the sample means is 8.5 – 7.5 = +1 inch. This means the average for Corn-e-stats minus the average for Stats-o-sweet is positive, making Corn-e-stats the larger of the two varieties, in terms of this sample. Is that difference enough to generalize to the entire population, though? That’s what this confidence interval is going to help you decide. Using the rest of the information you are given, find the confidence interval for the difference in mean cob length for the two brands: Your 95% confidence interval for the difference between the average lengths for these two varieties of sweet corn is 1 inch, plus or minus 0.9273 inches. (The lower end of the interval is 1 – 0.9273 = 0. 0727 inches; the upper end is 1 + 0. 9273 = 1. 9273 inches.) Notice all the values in this interval are positive. That means Corn-e-stats is estimated to be longer than Stats-o-sweet, based on your data. The temptation is to say, “Well, I knew Corn-e-stats corn was longer because its sample mean was 8.5 inches and Stat-o-sweet was only 7.5 inches on average. Why do I even need a confidence interval?” All those two numbers tell you is something about those 30 ears of corn sampled. You also need to factor in variation using the margin of error to be able to say something about the entire populations of corn.

Article / Updated 09-22-2022

A matrix is a rectangular array of numbers. Each row has the same number of elements, and each column has the same number of elements. Matrices can be classified as: square, identity, zero, column, and so on. Where did matrices come from? For most of their history, they were called arrays. There are references to arrays in Chinese, French, Italian, and many other mathematical works going back many hundreds of years. American mathematician George Dantzig's work with matrices during World War II allowed for the coordination of shipments of supplies and troops to various locations. Matrices are here to stay. You may be familiar with a method used to solve systems of linear equations using matrices, but this application just scratches the surface of what matrices can do. First, just in case you're not familiar with solving equations using matrices, let me give just a quick description. If you want to solve the following system of equations: You write the matrix: And then you perform row operations until you get the matrix: From that matrix, you know that the solution of the system of equations is x = 1, y = -3, and z = -5. Pretty slick, don't you think? But uses for matrices don't stop there. You can solve traffic control problems, transportation logistics problems (how much of each item to send to various distribution centers), dietary problems (how much of each food product is needed to meet several different dietary requirements), and so on. Matrices work well in graphing calculators and computer spreadsheets — just set up the problem and let the technology do all the work.

Article / Updated 09-19-2022

Your first step in any problem that involves partial fractions is to recognize which case you’re dealing with so that you can solve the problem. One case where you can use partial fractions is with repeated linear factors. These are difficult to work with because each factor requires more than one partial fraction. For each squared linear factor in the denominator, add two partial fractions in the following form: For each quadratic factor in the denominator that’s raised to the third power, add three partial fractions in the following form: Generally speaking, when a linear factor is raised to the nth power, add n partial fractions. For example, suppose that you want to integrate the following expression: This expression contains all linear factors, but one of these factors (x + 5) is nonrepeating and the other (x – 1) is raised to the third power. Set up your partial fractions this way: Which will yield: As you can see, this example adds one partial fraction to account for the nonrepeating factor and three to account for the repeating factor. When you start out with a linear factor, using partial fractions leaves you with an integral in the following form: Integrate all these cases by using the variable substitution u = ax + b so that du = a dx and This substitution results in the following integral: Here are a few examples:

Article / Updated 09-19-2022

Your first step in any problem that involves partial fractions is to recognize which case you’re dealing with so that you can solve the problem. One case where you can use partial fractions is when the denominator is the product of distinct quadratic factors — that is, quadratic factors that are nonrepeating. For each distinct quadratic factor in the denominator, add a partial fraction of the following form: For example, suppose that you want to integrate this function: The first factor in the denominator is linear, but the second is quadratic and can’t be decomposed to linear factors. So set up your partial fractions as follows: The number of distinct quadratic factors in the denominator tells you how many partial fractions you get. So in this example, two factors in the denominator yield two partial fractions. Working systematically with a system of equations Setting up a system of equations is an alternative method for finding the value of unknowns when you’re working with partial fractions. It’s not as simple as plugging in the roots of factors, but it’s your only option when the root of a quadratic factor is imaginary. Here’s a problem to illustrate this method: To start out, see how far you can get by plugging in the roots of equations. Begin by getting a common denominator on the right side of the equation: Now multiply the whole equation by the denominator: 5x – 6 = (A)(x2 + 3) + (Bx + C)(x – 2) The root of x – 2 is 2, so let x = 2 and see what you get: Now you can substitute Unfortunately, x2 + 3 has no root in the real numbers, so you need a different approach. First, get rid of the parentheses on the right side of the equation: Next, combine similar terms (using x as the variable by which you judge similarity). This is just algebra: Because this equation works for all values of x, you now take what appears to be a questionable step, breaking this equation into three separate equations as follows: At this point, a little algebra tells you that So you can substitute the values of A, B, and C back into the partial fractions: You can simplify the second fraction a bit: Quadratic factors of the form (ax2 + C) When you start out with a quadratic factor of the form (ax2 + C), using partial fractions results in the following two integrals: Integrate the first by using the variable substitution u = ax2 + C so that du = 2ax dx and This substitution results in the following integral: Here are some examples: To evaluate the second integral, use the following formula: Quadratic factors of the form (ax2 + bx + C) Most math teachers have at least a shred of mercy in their hearts, so they don’t tend to give you problems that include this most difficult case. When you start out with a quadratic factor of the form (ax2 + bx + C), using partial fractions results in the following integral: Okay, that’s way too many letters and not nearly enough numbers. Here’s an example: This is about the hairiest integral you’re ever going to see at the far end of a partial fraction. To evaluate it, you want to use the variable substitution u = x2 + 6x + 13 so that du = (2x + 6) dx. If the numerator were 2x + 6, you’d be in great shape. So you need to tweak the numerator a bit. First multiply it by 2 and divide the whole integral by 2: Because you multiplied the entire integral by 1, no net change has occurred. Now add 6 and –6 to the numerator: You have added 0 to the integral, which didn’t change its value. At this point, you can split the integral in two: At this point, you can use variable substitution to change the first integral as follows: To solve the second integral, complete the square in the denominator: Divide the b term (6) by 2 and square it, and then represent the C term (13) as the sum of this and whatever’s left: Now split the denominator into two squares: To evaluate this integral, use formula shown in the previous section: So here’s the final answer for the second integral: Therefore, piece together the complete answer as follows:

Article / Updated 09-19-2022

Your first step in any problem that involves partial fractions is to recognize which case you’re dealing with so that you can solve the problem. One case where you can use partial fractions is with repeated quadratic factors. This is your worst nightmare when it comes to partial fractions, because the denominator includes repeated quadratic factors. For each squared quadratic factor in the denominator, add two partial fractions in the following form: For each quadratic factor in the denominator that’s raised to the third power, add three partial fractions in the following form: Generally speaking, when a quadratic factor is raised to the nth power, add n partial fractions. For example: This denominator has one nonrepeating linear factor (x – 8), one nonrepeating quadratic factor (x2 + x + 1), and one quadratic expression that’s squared (x2 + 3). Here’s how you set up the partial fractions: This example adds one partial fraction for each of the nonrepeating factors and two partial fractions for the squared factor. When you start out with a quadratic factor of the form (ax2 + C), using partial fractions results in the following two integrals: Integrate the first by using the variable substitution u = ax2 + C so that du = 2ax dx and This substitution results in the following integral: Here are some examples: To evaluate the second integral, use the following formula: Most math teachers have at least a shred of mercy in their hearts, so they don’t tend to give you problems that include this most difficult case. When you start out with a quadratic factor of the form (ax2 + bx + C), using partial fractions results in the following integral: Okay, that’s way too many letters and not nearly enough numbers. Here’s an example: This is about the hairiest integral you’re ever going to see at the far end of a partial fraction. To evaluate it, you want to use the variable substitution u = x2 + 6x + 13 so that du = (2x + 6) dx. If the numerator were 2x + 6, you’d be in great shape. So you need to tweak the numerator a bit. First multiply it by 2 and divide the whole integral by 2: Because you multiplied the entire integral by 1, no net change has occurred. Now add 6 and –6 to the numerator: This time, you add 0 to the integral, which doesn’t change its value. At this point, you can split the integral in two: At this point, you can use variable substitution to change the first integral as follows: To solve the second integral, complete the square in the denominator: Divide the b term (6) by 2 and square it, and then represent the C term (13) as the sum of this and whatever’s left: Now split the denominator into two squares: To evaluate this integral, use the same formula from the previous section: So here’s the final answer for the second integral: Therefore, piece together the complete answer as follows:

Article / Updated 09-19-2022

You can find the average value of a function over a closed interval by using the mean value theorem for integrals. The best way to understand the mean value theorem for integrals is with a diagram — look at the following figure. The graph on the left shows a rectangle whose area is clearly less than the area under the curve between 2 and 5. This rectangle has a height equal to the lowest point on the curve in the interval from 2 to 5. The middle graph shows a rectangle whose height equals the highest point on the curve. Its area is clearly greater than the area under the curve. By now you’re thinking, “Isn’t there a rectangle taller than the short one and shorter than the tall one whose area is the same as the area under the curve?” Of course. And this rectangle obviously crosses the curve somewhere in the interval. This so-called mean value rectangle, shown on the right, basically sums up the Mean Value Theorem for Integrals. It’s really just common sense. But here’s the mumbo jumbo. The mean value theorem for integrals: If f (x) is a continuous function on the closed interval [a, b], then there exists a number c in the closed interval such that The theorem basically just guarantees the existence of the mean value rectangle. The area of the mean value rectangle — which is the same as the area under the curve — equals length times width, or base times height, right? This height is the average value of the function over the interval in question. Here’s an example. What’s the average speed of a car between t = 9 seconds and t = 16 seconds whose speed in feet per second is given by the function, According to the definition of average value, this average speed is given by Determine the area under the curve between 9 and 16. This area, by the way, is the total distance traveled from 9 to 16 seconds. Do you see why? Consider the mean value rectangle for this problem. Its height is a speed (because the function values, or heights, are speeds) and its base is an amount of time, so its area is speed times time which equals distance. Alternatively, recall that the derivative of position is velocity. So, the antiderivative of velocity — what you just did in this step — is position, and the change of position from 9 to 16 seconds gives the total distance traveled. Divide this area, total distance, by the time interval from 9 to 16, namely 7. ≈ 105.7 feet per second It makes more sense to think about these problems in terms of division: area equals base times height, so the height of the mean value rectangle equals its area divided by its base.

Article / Updated 09-16-2022

Using geometry symbols will save time and space when writing proofs, properties, and figuring formulas. The most commonly used geometry symbols and their meanings are shown below.