How to of the Day

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• How to Escape from Being Tied Up
• How to Befriend a Wild Cat or Kitten
• How to Integrate

How to Escape from Being Tied Up Posted: 20 Mar 2017 05:00 PM PDT Most of us go through our day-to-day lives experiencing little or no personal danger. However, in the unlikely case that you do find yourself in a threatening situation, it's smart to know how to respond and defend yourself ahead of time. Should you find yourself attacked and tied up, whether from military action, a kidnapping while outside of your home or traveling abroad, or aggressive home invaders, you'll need to know how to escape from your bonds. It's important to not panic when you're tied, but focus on loosening or cutting free from your bonds. EditSteps EditEscaping When You're Tied with Rope Position yourself to avoid being tied tightly. The more loosely you're bound, the easier it will be for you to escape. Resist letting your captors tie you as tightly as possible. For example, if your wrists are being tied in front of you, hold your knuckles from both hands together, and pull your hands in towards your chest. Although it looks like you're cooperating, this gesture will create a gap between your wrists.[1] Alternatively, you can present your hands to be tied with both hands in front of you, crossed at the wrist. Rotate your lower wrist about 45 degrees while you're being tied, so that, once the ropes are tight, you can straighten out your wrist and loosen the rope.[2] Tense all of your muscles while you are being tied up. If more than just your wrists or ankles are being bound, this will help you escape. When your muscles are tensed, they expand and take up more room than when they're relaxed. This will make your body larger, while your captors are tying you tightly. Consequently, when you relax your muscles, your body will shrink slightly and the ropes will be loser, giving your more room to move and eventually escape. This is a technique used by most escape artists and allows the ropes to be slightly looser when you relax you muscles. If captors are tying a rope around your chest, take a deep breath and expand your lungs as much as possible.[3] If you can make a loose spot big enough to do so, slip out of it. Wriggle out of the ropes around your hands. Once your captors have turned their backs or left the room you're in, rotate your wrists back and forth to loosen them. You can also use your teeth to pull on one strand of the rope to make it looser.[4] As the ropes continue to loosen, you may be able to wriggle out of them. If your hands are tied to your torso or sides, wriggle your hands until they're at a narrower part of your body (such as directly in front of you). The ropes will be looser here, and you'll be able to slip out.[5] If your arms are pinned to your stomach, chest or torso, bend one arm upwards and try lifting the bonds. If the rope is at all lose, you may be able to slip it off over your head. Try to cut the ropes around your wrists. You need your hands to untie any other ropes, so always begin by freeing your hands. Ropes (and also phone and electrical cords) can be cut through with friction, so you'll need to find a hard object to rub the binding ropes against.[6] Look for surfaces like an exposed corner of a cement wall, a table edge, or a granite countertop. If you're alone in a room, look around for a sharp object, such as a knife, scissor blade, etc. If you use one of these objects to cut the rope while your hands are tied, take care not to cut or injure yourself. If you have a key or small knife in your pocket, try to get it out without anyone noticing. If you can cut the ropes quickly, you'll be well on your way to escaping. Kick off your shoes before you free your feet and legs. If your hands are impossible to free, you may need to free your feet first. First kick of your shoes, as it will be easier to slip the ropes if you are just wearing socks. First see if the rope is loose enough to wriggle out of—if it's not, bend over and try untying the knot with your teeth. Once your feet are free, use your tied hands to pull the leg bonds down until they have gone off over your feet. Plan your moment of escape carefully. Don't plan to run out the door as soon as you're free; you'll need to make a tactical exit. You should bide your time and pick the best possible moment to escape. When your captors have turned their backs or left the room, make your break.[7] Get far away as quickly as you can. If you expect to be pursued by your captors, try to blend in or hide in your environment, and arm yourself (for example, with a piece of rebar) if needed. Also, you can gather information about your captors that could help the police track them down. Pay attention to their looks and physical appearance, any tattoos and scars, and the sound of their voices. EditBreaking Free from Zip Ties Break the locking mechanism on the zip ties. This is the weakest point on the zip tie, and will be the easiest to break. To break the ties, form your hands into fists with the knuckles pressed together Raise your bound hands above your head, then bring them down sharply. At the same time, pull your elbows apart and press your wrists hard into your abdomen.[8] This should exert enough pressure to snap the lock mechanism on the zip ties. If your hands are bound in front of you, tighten the zip tie as much as possible, as it will be more fragile when fully taut. A looser zip tie will be harder to break free from. Cut through the zip ties with friction. If you can maneuver yourself to a hard surface, rub the band of the zip ties over the surface repeatedly to build up friction and, eventually, cut through the zip ties.[9] Paracord or Kevlar string are heat resistant and can be used to create enough friction to cut through zip ties or rope. If you're concerned you may be tied up, or traveling in a dangerous area, consider replacing your shoelaces with paracord or Kevlar string.[10] To escape, tie your shoelaces from both feet together, with the knot between your bound wrists. Then use a "bicycle" motion to create friction and cut your zip ties off. Slip out from the zip ties. When you're being tied, clench your fists tightly to expand the muscles in your wrists. This will make your wrists larger, and make your bonds looser once you've relaxed your hands.[11] If you do this correctly, you should be able to wriggle your hands out of the zip ties, without injuring yourself. If you've been tied tightly, rotate your hands and move your wrists back and forth against each other. This may loosen the plastic zip cords, and create enough room for you to slip your hands free.[12] This may take time, so make sure your captors don't observe you trying to escape. EditGetting Out of Duct Tape Break the tape in front of you. Although duct tape is incredibly adhesive, it is still vulnerable to tears and breaks. If your hands are bound in front of you, raise them above your head, and then bring them down quickly against your abdomen, pulling your elbows apart at the same time.[13] Unlike zip ties, breaking duct tape in this method is unlikely to cut your wrists. Chew through the duct tape. Since duct tape is not quite as sturdy as rope, it can be bent and torn more easily. Use this to your advantage. You may be able to chew through it, or use your teeth to tear the tape, and then pry the tape apart.[14] If you can't break the tape with your teeth, try to peel it away from your skin with your teeth or mouth. This will give your more space to wriggle out of the tape. Get the duct tape wet in order to loosen it. Like any other type of tape, duct tape loses nearly all of its adhesive quality when it is wet. If you are near water bottles or a natural water source (even if there's water on the floor or in a sink of the room where you're tied), get to the water and moisten the tape.[15] After a period of time, the tape will loosen and you'll be able to pull your hands or feet out. If you can lick or spit on the duct tape, even saliva from your mouth will help loosen the adhesion. EditTips Try practicing to escape by asking a friend or sibling to tie you up. Use the techniques described to escape, but if you can't do it the first time, that's fine—it takes practice. Make sure you use a rope that is OK to cut, if you can't slip the rope off. If you're escaping from a dangerous situation, make sure your captor doesn't notice your escape—otherwise they may re-tie you, and more effectively. If your hands are tied behind your back and you're in a standing position. Try to bring your hands close to the ground and hop over them, which will bring them in front of you. This will make it easier to loosen them. EditWarnings Immediately call the police after escaping. It may be painful when you escape from being tied up, especially if you've been bound with zip ties or tight rope. Don't let this distract you. You can seek medical attention later if you need it, but your top priority is escaping. EditRelated wikiHows Thwart an Abduction Attempt Chase Somebody on Foot Get Away from an Attacker Stand up for Yourself Tie Yourself up With Rope Hogtie Someone EditSources and Citations Cite error: <ref> tags exist, but no <references/> tag was found

How to Befriend a Wild Cat or Kitten Posted: 20 Mar 2017 09:00 AM PDT Many cities are full of stray cats and it is estimated that there are over 70 million stray cats in the US alone.[1] Many municipalities are unable or unwilling to take any action, beyond capturing a few cats to euthanize them, due to a lack of time, money, and interest. Because there is little help to stray cats and their risk of death from injuries, disease and poor nutrition, you may feel compelled to help a wild cat or kitten yourself. This process may take some time, so you need to be patient, but it can be very fulfilling to befriend and help a cat in need. EditSteps EditLuring a Wild Cat to You Differentiate between a stray cat and a feral cat. A stray cat is someone's former cat that no longer has a home; a feral cat was born in the wild and is a cat that is wary of humans and generally isn't adoptable due to not being adapted to humans.[2] You may have success at befriending a stray cat; a feral cat probably not although it may become acclimated to your presence. Whether you have a stray or a feral cat around your home, capturing (or trapping) the cat and bringing it to the veterinarian for vaccinations and to be spayed or neutered is an excellent civic duty. Feral cats that are "fixed" can be released back where you found them.[3] You'll have the satisfaction of knowing they won't be able to reproduce and contribute to the feral cat population. Take safety precautions. Wild cats can be unpredictable so you will need to practice some safety precautions as you are trying to befriend it. Cat bites usually become infected, some very seriously, so you will need to wear long sleeves and pants when trying to befriend the cat. Another concern is rabies in a cat of unknown vaccination status. Use caution and common sense. If the cat starts to hiss or growl, looks sick (runny nose or eyes, scabs, sneezes/coughs, breathes heavily), or acts bizarrely don't try to befriend it. Call animal control and retreat to your home. Begin befriending the cat. If the cat appears healthy and isn't acting defensive and mean towards you, you can try befriending it. Find a place where the cat usually stays, and wait nearby until the cat reaches its spot. Sit down or lie down, if you can, or at least crouch. The cat will find you much less intimidating that way. Stay there for a while. Teach the cat you won't hurt it. This should be a fair distance away, 10 feet or so, so you don't intimidate the cat. Offer the cat food. Try leaving out some smelly cat food (wet) or even a can of tuna to attract the cat while you are waiting for it to arrive. You want to get the cat to associate you with something good, namely a free lunch. You only need to leave out smelly food the first day. After this leave out a little dry kibble to keep the cat coming back for more. Continue putting little bits of tuna closer and closer to you. Each day the cat comes to eat, move your spot another foot closer. If you hold the cat food in your hand for a while, your scent will be on it, too. This won't discourage the cat from eating it, but it may cause the cat to associate you with food, which is good in general. Put out your hand to let the cat sniff it. If the cat starts hissing, or its ears go back, etc., you're too close. Pull your hand back slightly and slowly. Show the cat some affection. Eventually the cat should get close to you then put out your hand let the cat sniff your hand. Continue feeding the cat, and sitting nearby quietly, and eventually you should earn its trust to the point it will come up to you for food and gentle pats. Don't expect this to happen right away. Don't expect them to even eat any of the tuna right away. EditTaking Care of a Wild Cat or Kitten Help find the cat's owner. Ask around your neighborhood to see if anyone has lost their cat. Cats that get outside can get lost or roam around. Make some calls to local veterinary clinics and pet stores to see if anyone is missing a cat; taking a photo with your phone and making up fliers or posting to social media sites can also help reunite a stray cat with its owners. If the cat is feral instead of stray, it will not have an owner to be returned to. Get the cat checked out by a veterinarian. Once you are able to approach the cat and pet it, make an appointment with your veterinarian to have the cat examined and neutered or spayed. If funds are tight, call your local humane society to see if there are any funds available to assist with these costs. The cat will have a tiny bit of blood drawn to test for the feline leukemia virus. If the results are positive your veterinarian will discuss options with you, which include isolating the cat from other cats for its life or euthanasia. If it is negative, a fecal sample will be looked at to determine if the cat has intestinal parasites and it will be given the appropriate medication. In addition, the veterinarian will check for fleas and ticks and treat if needed. Vaccines (rabies, distemper, and possibly feline leukemia) will be given and its sterilization operation will be performed. A microchip may also be placed under the skin if so desired. This is highly recommended. Find a home for the cat or kitten. Now you will have a healthy pet to take home after its operation and vaccinations. Your hard work will pay off knowing you have saved another cat from a harsh life on the street. Either adopt the cat yourself or find another loving home for it to go to. EditTips If the cat comes into your ownership, you may want to consider buying toys for it. Squeaky toys, scratching post, or even a ball of yarn can keep it occupied for those times when you are too busy to play with it. If the cat seems very sleek and well-fed, do your best to make sure it doesn't already have an owner. If the cat is truly afraid of you, leave it alone. Feel free to leave food for it; it may eventually warm up to you. If you notice the cat giving you a slow-blink, make sure to slow-blink back. This means 'Love you' in cat, and it may tell you that a cat has fully accepted you and is ready to be adopted. However, if it is giving you an unblinking stare, back up and avert eye contact. This means 'Get out of my territory. If the cat is just a baby or a kitten, there may be a litter near by. Pay close attention! EditWarnings Remember, only adopt a cat if you're willing to take care of it for its whole life, or find it another home. If you need to find it another home, it's best to find someone to adopt it, not to take it to a pound. These tend to be overrun by animals, especially cats, and may be forced to put down cats if they don't have any more room. So even if they don't specifically put down your cat, you may cause another one to be put down. Remember always to wash your hands thoroughly after touching the cat until you've had a chance to bring it to the veterinarian. Refrain from posting about the cat in question online. It may be claimed by its owner that way but it can also be claimed by poachers, hoarders, people who use cats in dog rings, and people who sell cats to be tested on. If you really want to put an add online don't post a picture of it or describe it. The true owner should be able to describe it to you. Never run towards any cat. It will find this as aggressive behavior and it may try to bite or scratch you. If the cat is aggressive, don't try to befriend it. Even if you manage, it will still be a wild cat, and may be nasty. EditRelated wikiHows Pet a High Strung Cat Litter Train a Cat Keep a Cat from Running Away when It Is Moved Choose the Right Cat Insurance Act Like a Cat Bathe a Cat Bathe Your Cat With a Damp Towel Train a Kitten Earn the Trust of a Feral Kitten Care for Feral Cats Care for Siamese Kittens EditSources and Citations Cite error: <ref> tags exist, but no <references/> tag was found

How to Integrate Posted: 20 Mar 2017 01:00 AM PDT Integration is the inverse operation of differentiation. It is commonly said that differentiation is a science, while integration is an art. The reason is because integration is simply a harder task to do - while a derivative is only concerned with the behavior of a function at a point, an integral, being a glorified sum, requires global knowledge of the function. So while there are some functions whose integrals can be evaluated using the standard techniques in this article, many more cannot. We go over the basic techniques of single-variable integration in this article and apply them to functions with antiderivatives. EditSteps EditThe Basics Understand the notation for integration. An integral ${\displaystyle \int _{a}^{b}f(x){\mathrm {d} }x}$ consists of four parts. The ${\displaystyle \int }$ is the symbol for integration. It is actually an elongated S. The function ${\displaystyle f(x)}$ is called the integrand when it is inside the integral. The differential ${\displaystyle {\mathrm {d} }x}$ intuitively is saying what variable you are integrating with respect to. Because (Riemann) integration is just a sum of infinitesimally thin rectangles with a height of ${\displaystyle f(x),}$ we see that ${\displaystyle {\mathrm {d} }x}$ refers to the width of those rectangles. The letters ${\displaystyle a}$ and ${\displaystyle b}$ are the boundaries. An integral does not need to have boundaries. When this is the case, we say that we are dealing with an indefinite integral. If it does, then we are dealing with a definite integral. Throughout this article, we will go over the process of finding antiderivatives of a function. An antiderivative is a function whose derivative is the original function we started with. Understand the definition of an integral. When we talk about integrals, we usually refer to Riemann integrals; in other words, summing up rectangles. Given a function ${\displaystyle f(x),}$ a rectangle width of ${\displaystyle \Delta x,}$ and an interval ${\displaystyle [a,b],}$ the area of the first rectangle is given by ${\displaystyle f(x_{1})\Delta x_{1},}$ because it is just the base times the height (the value of the function). Similarly, the area of the second rectangle is ${\displaystyle f(x_{2})\Delta x_{2}.}$ Generalizing, we say the area of the ith rectangle is ${\displaystyle f(x_{i})\Delta x_{i}.}$ In summation notation, this can be represented in the following manner. ${\displaystyle \sum _{i=1}^{n}f(x_{i})\Delta x_{i}}$ If this is the first time you have seen a summation symbol, it may look scary...but it's not complicated at all. All this is saying is that we are summing up the area of ${\displaystyle n}$ rectangles. (The variable ${\displaystyle i}$ is known as a dummy index.) However, as you can guess, the area of all the rectangles is bound to be slightly different from the true area. We solve this by sending the number of rectangles to infinity. As we increase the number of rectangles, the area of all the rectangles better approximates the area under the curve. That's what the diagram above shows (see the tips for what the graph in the middle shows). The limit as ${\displaystyle n\to \infty }$ is what we define as the integral of the function ${\displaystyle f(x)}$ from ${\displaystyle a}$ to ${\displaystyle b.}$ ${\displaystyle \int _{a}^{b}f(x){\mathrm {d} }x=\lim _{n\to \infty }\sum _{i=1}^{n}f(x_{i})\Delta x_{i}}$ Of course, this limit has to exist in order for the integral to have any meaning. If such a limit does not exist on the interval, then we say that ${\displaystyle f(x)}$ does not have an integral over the interval ${\displaystyle [a,b].}$ In this article (and in almost every physical application), we only deal with functions where these integrals exist. Remember ${\displaystyle +C}$ when evaluating indefinite integrals! One of the most common mistakes people can make is forgetting to add the constant of integration. The reason why this is needed is because antiderivatives are not unique. In fact, a function can have an infinite number of antiderivatives. They are allowed because the derivative of a constant is 0. EditPower Rule Consider a monomial ${\displaystyle x^{n}}$. Perform the power rule for integrals. This is the same power rule for derivatives, but in reverse. We increase the power by 1, and divide by the new power. Don't forget to add the constant of integration ${\displaystyle C.}$ ${\displaystyle \int x^{n}{\mathrm {d} }x={\frac {x^{n+1}}{n+1}}+C}$ To verify that this power rule holds, differentiate the antiderivative to recover the original function. The power rule holds for all functions of this form with degree ${\displaystyle n}$ except when ${\displaystyle n=-1.}$ We will see why later. Apply linearity. Integration is a linear operator, which means that the integral of a sum is the sum of the integrals, and the coefficient of each term can be factored out, like so: ${\displaystyle \int (ax^{n}+bx^{m}){\mathrm {d} }x=a\int x^{n}{\mathrm {d} }x+b\int x^{m}{\mathrm {d} }x}$ This should be familiar because the derivative is also a linear operator; the derivative of a sum is the sum of the derivatives. Linearity does not apply just for integrals of polynomials. It applies to any integral where the integrand is a sum of two or more terms. Find the antiderivative of the function ${\displaystyle f(x)=x^{4}+2x^{3}-5x^{2}-1}$. This is a polynomial, so using the property of linearity and the power rule, the antiderivative can easily be computed. To find the antiderivative of a constant, remember that ${\displaystyle x^{0}=1,}$ so the constant is really just the coefficient of ${\displaystyle x^{0}.}$ ${\displaystyle \int (x^{4}+2x^{3}-5x^{2}-1){\mathrm {d} }x={\frac {1}{5}}x^{5}+{\frac {2}{4}}x^{4}-{\frac {5}{3}}x^{3}-x+C}$ Find the antiderivative of the function ${\displaystyle f(x)={\frac {2x^{2}+3x-1}{x^{1/3}}}}$. This may seem like a function that defies our rules, but a moment's glance reveals that we can separate the fraction into three fractions and apply linearity and the power rule to find the antiderivative. ${\displaystyle {\begin{aligned}\int f(x){\mathrm {d} }x&=\int {\frac {2x^{2}+3x-1}{x^{1/3}}}{\mathrm {d} }x\\&=\int \left({\frac {2x^{2}}{x^{1/3}}}+{\frac {3x}{x^{1/3}}}-{\frac {1}{x^{1/3}}}\right){\mathrm {d} }x\\&=\int (2x^{5/3}+3x^{2/3}-x^{-1/3}){\mathrm {d} }x\\&=2\cdot {\frac {3}{8}}x^{8/3}+3\cdot {\frac {3}{5}}x^{5/3}-{\frac {3}{2}}x^{2/3}+C\\&={\frac {3}{4}}x^{8/3}+{\frac {9}{5}}x^{5/3}-{\frac {3}{2}}x^{2/3}+C\end{aligned}}}$ The common theme is that you must perform whatever manipulations in order to get the integral into a polynomial. From there, integration is easy. Judging whether the integral is easy enough to brute-force, or requires some algebraic manipulation first, is where the skill lies. EditDefinite Integration Consider the integral below. Unlike the integration process in part 2, we also have bounds to evaluate at. ${\displaystyle \int _{2}^{3}x^{2}{\mathrm {d} }x}$ Use the fundamental theorem of calculus. This theorem is in two parts. The first part was stated in the first sentence of this article: integration is the inverse operation of differentiation, so integrating and then differentiating a function recovers the original function. The second part is stated below. Let ${\displaystyle F(x)}$ be an antiderivative of ${\displaystyle f(x).}$ Then ${\displaystyle \int _{a}^{b}f(x){\mathrm {d} }x=F(b)-F(a).}$ This theorem is incredibly useful because it simplifies the integral and means that the definite integral is completely determined by just the values on its boundaries. There is no need to sum up rectangles anymore to compute integrals. All we need to do now is to find antiderivatives, and evaluate at the bounds! Evaluate the integral stated in step 1. Now that we have the fundamental theorem as a tool for solving integrals, we can easily calculate the value of the integral as defined above. ${\displaystyle {\begin{aligned}\int _{2}^{3}x^{2}{\mathrm {d} }x&={\frac {1}{3}}x^{3}{\Bigg |}_{2}^{3}\\&={\frac {1}{3}}(3)^{3}-{\frac {1}{3}}(2)^{3}\\&={\frac {19}{3}}\end{aligned}}}$ Again, the fundamental theorem of calculus does not just apply to functions like ${\displaystyle f(x)=x^{2}.}$ The fundamental theorem can be used to integrate any function, as long as you can find an antiderivative. Evaluate the integral with the boundaries swapped. Let's see what happens here. ${\displaystyle {\begin{aligned}\int _{3}^{2}x^{2}{\mathrm {d} }x&={\frac {1}{3}}x^{3}{\Bigg |}_{3}^{2}\\&={\frac {1}{3}}(2)^{3}-{\frac {1}{3}}(3)^{3}\\&=-{\frac {19}{3}}\end{aligned}}}$ We just obtained the negative of the answer we got before. This illustrates an important property of definite integrals. Swapping the boundaries negates the integral. ${\displaystyle \int _{b}^{a}f(x){\mathrm {d} }x=-\int _{a}^{b}f(x){\mathrm {d} }x}$ EditAntiderivatives of Common Functions Memorize the antiderivatives of exponential functions. In the following steps, we list commonly encountered functions like the exponential and trigonometric functions. All are widely encountered, so knowing what their antiderivatives are is crucial to building up integrating skills. Remember that indefinite integrals have an extra ${\displaystyle C,}$ because the derivative of a constant is 0. ${\displaystyle \int e^{x}{\mathrm {d} }x=e^{x}+C}$ ${\displaystyle \int a^{x}{\mathrm {d} }x={\frac {a^{x}}{\ln a}}+C}$ Memorize the antiderivatives of trigonometric functions. These are just the derivatives applied backwards and should be familiar. The sines and cosines are encountered far more often and should definitely be memorized. Hyperbolic analogues are similarly found, though they are encountered less often. ${\displaystyle \int \sin x{\mathrm {d} }x=-\cos x+C}$ ${\displaystyle \int \cos x{\mathrm {d} }x=\sin x+C}$ ${\displaystyle \int \sec ^{2}x{\mathrm {d} }x=\tan x+C}$ ${\displaystyle \int \csc ^{2}x{\mathrm {d} }x=-\cot x+C}$ ${\displaystyle \int \sec x\tan x{\mathrm {d} }x=\sec x+C}$ ${\displaystyle \int \csc x\cot x{\mathrm {d} }x=-\csc x+C}$ Memorize the antiderivatives of inverse trigonometric functions. These should not really be considered an exercise in "memorization." As long as you are familiar with the derivatives, then most of these antiderivatives should be familiar as well. ${\displaystyle \int {\frac {1}{\sqrt {1-x^{2}}}}{\mathrm {d} }x=\sin ^{-1}x+C}$ ${\displaystyle \int {\frac {-1}{\sqrt {1-x^{2}}}}{\mathrm {d} }x=\cos ^{-1}x+C}$ ${\displaystyle \int {\frac {1}{1+x^{2}}}{\mathrm {d} }x=\tan ^{-1}x+C}$ Memorize the antiderivative of the reciprocal function. Previously, we said that the function ${\displaystyle f(x)=x^{-1},}$ or ${\displaystyle f(x)={\frac {1}{x}},}$ was an exception to the power rule. The reason is because the antiderivative of this function is the logarithmic function. ${\displaystyle \int {\frac {1}{x}}{\mathrm {d} }x=\ln |x|+C}$ (Sometimes, authors like to put the ${\displaystyle {\mathrm {d} }x}$ in the numerator of the fraction, so it reads like ${\displaystyle \int {\frac {{\mathrm {d} }x}{x}}.}$ Be aware of this notation.) The reason for the absolute value in the logarithm function is subtle, and requires a more thorough understanding of real analysis in order to fully answer. For now, we will just live with the fact that the domains become the same when the absolute value bars are added. Evaluate the following integral over the given bounds. Our function is given as ${\displaystyle f(x)=2\cos x+\tan ^{2}x-6.}$ Here, we don't know the antiderivative of ${\displaystyle \tan ^{2}x,}$ but we can use a trigonometric identity to rewrite the integrand in terms of a function which we know the antiderivative of - namely, ${\displaystyle 1+\tan ^{2}x=\sec ^{2}x.}$ ${\displaystyle {\begin{aligned}\int _{\pi /4}^{\pi /3}f(x){\mathrm {d} }x&=\int _{\pi /4}^{\pi /3}(2\cos x+\tan ^{2}x-6){\mathrm {d} }x\\&=\int _{\pi /4}^{\pi /3}(2\cos x+\sec ^{2}x-7){\mathrm {d} }x\\&=(2\sin x+\tan x-7x){\Big |}_{\pi /4}^{\pi /3}\\&=\left(2\sin {\frac {\pi }{3}}+\tan {\frac {\pi }{3}}-{\frac {7\pi }{3}}\right)-\left(2\sin {\frac {\pi }{4}}+\tan {\frac {\pi }{4}}-{\frac {7\pi }{4}}\right)\\&=2{\sqrt {3}}-{\sqrt {2}}-1-{\frac {7\pi }{12}}\end{aligned}}}$ If you need a decimal approximation, you can use a calculator. Here, ${\displaystyle 2{\sqrt {3}}-{\sqrt {2}}-1-{\frac {7\pi }{12}}\approx -0.7827.}$ EditIntegrals of Symmetric Functions Evaluate the integral of an even function. Even functions are functions with the property that ${\displaystyle f(-x)=f(x).}$ In other words, you should be able to replace every ${\displaystyle x}$ with a ${\displaystyle -x}$ and get the same function. An example of an even function is ${\displaystyle x^{2}.}$ Another example is the cosine function. All even functions are symmetric about the y-axis. ${\displaystyle \int _{-1}^{1}(\cos x+x^{4}){\mathrm {d} }x}$ Our integrand is even. We can immediately integrate by using the fundamental theorem of calculus, but if we look more carefully, we see that the bounds are symmetric about ${\displaystyle x=0.}$ That means that the integral from -1 to 0 is going to give us the same value as the integral from 0 to 1. So what we can do is we can change the bounds to 0 and 1 and factor out a 2. ${\displaystyle \int _{-1}^{1}(\cos x+x^{4}){\mathrm {d} }x=2\int _{0}^{1}(\cos x+x^{4}){\mathrm {d} }x}$ It might not seem like much to do this, but we will immediately see that our work is simplified. After finding the antiderivative, notice that we only need to evaluate it at ${\displaystyle x=1.}$ The antiderivative at ${\displaystyle x=0}$ will not contribute to the integral. ${\displaystyle 2\int _{0}^{1}(\cos x+x^{4}){\mathrm {d} }x=2\sin 1+{\frac {2}{5}}}$ In general, whenever you see an even function with symmetric boundaries, you should perform this simplification in order to make less arithmetic mistakes. ${\displaystyle \int _{-a}^{a}f(x){\mathrm {d} }x=2\int _{0}^{a}f(x){\mathrm {d} }x,\ \ f(x)\ {\mathrm {even} }}$ Evaluate the integral of an odd function. Odd functions are functions with the property that ${\displaystyle f(-x)=-f(x).}$ In other words, you should be able to replace every ${\displaystyle x}$ with a ${\displaystyle -x}$ and then get the negative of the original function. An example of an odd function is ${\displaystyle x^{3}.}$ The sine and tangent functions are also odd. All odd functions are symmetric about the origin (imagine rotating the negative part of the function by 180° - it will then stack on top of the positive part of the function). If the bounds are symmetric, then the integral will be 0. ${\displaystyle \int _{-\pi /2}^{\pi /2}(2x^{3}+2\sin x){\mathrm {d} }x}$ We could evaluate this integral directly...or we can recognize that our integrand is odd. Furthermore, the boundaries are symmetric about the origin. Therefore, our integral is 0. Why is this the case? It is because the antiderivative is even. Even functions have the property that ${\displaystyle f(-x)=f(x),}$ so when we evaluate at the bounds ${\displaystyle -a}$ and ${\displaystyle a,}$ then ${\displaystyle F(-a)=F(a)}$ immediately implies that ${\displaystyle F(a)-F(-a)=0.}$ ${\displaystyle \int _{-\pi /2}^{\pi /2}(2x^{3}+2\sin x){\mathrm {d} }x=0}$ The properties of these functions are very potent in simplifying the integrals, but the boundaries must be symmetric. Otherwise, we will need to evaluate the old way. ${\displaystyle \int _{-a}^{a}f(x){\mathrm {d} }x=0,\ \ f(x)\ {\mathrm {odd} }}$ EditU-Substitution See the main article on how to perform u-substitutions. U-substitution is a technique that changes variables with the hope of obtaining an easier integral. As we will see, it is the analogue of the chain rule for derivatives. Evaluate the integral of ${\displaystyle e^{ax}}$. What do we do when the exponent has a coefficient in it? We use u-substitution to change variables. It turns out that these kinds of u-subs are the easiest to perform, and they are done so often, the u-sub is often skipped. Nevertheless, we will show the entire process. ${\displaystyle \int e^{ax}{\mathrm {d} }x}$ Choose a ${\displaystyle u}$ and find ${\displaystyle {\mathrm {d} }u}$. We choose ${\displaystyle u=ax}$ so that we get a ${\displaystyle e^{u}}$ in the integrand, a function whose antiderivative we are familiar with - itself. Then we must replace ${\displaystyle {\mathrm {d} }x}$ with ${\displaystyle {\mathrm {d} }u,}$ but we need to make sure that we are keeping track of our terms. In this example, ${\displaystyle {\mathrm {d} }u=a{\mathrm {d} }x,}$ so we need to divide the whole integral by ${\displaystyle a}$ to compensate. ${\displaystyle \int e^{ax}{\mathrm {d} }x={\frac {1}{a}}\int e^{u}{\mathrm {d} }u}$ Evaluate and rewrite in terms of the original variable. For indefinite integrals, you must rewrite in terms of the original variable. ${\displaystyle \int e^{ax}{\mathrm {d} }x={\frac {1}{a}}e^{ax}+C}$ Evaluate the following integral with the given boundaries. This is a definite integral, so we need to evaluate the antiderivative at the boundaries. We will also see that this u-sub is a case where you need to "back-substitute." ${\displaystyle \int _{0}^{1}x{\sqrt {2x+3}}{\mathrm {d} }x}$ Choose a ${\displaystyle u}$ and find ${\displaystyle {\mathrm {d} }u}$. Make sure to change your boundaries as well according to your substitution. We choose ${\displaystyle u=2x+3}$ so that we simplify the square root. Then ${\displaystyle {\mathrm {d} }u=2{\mathrm {d} }x,}$ and the bounds then go from 3 to 5. However, after replacing the ${\displaystyle {\mathrm {d} }x}$ with a ${\displaystyle {\mathrm {d} }u,}$ we still have an ${\displaystyle x}$ in the integrand. ${\displaystyle \int _{0}^{1}x{\sqrt {2x+3}}{\mathrm {d} }x={\frac {1}{2}}\int _{3}^{5}x{\sqrt {u}}{\mathrm {d} }u}$ Solve for ${\displaystyle x}$ in terms of ${\displaystyle u}$ and substitute. This is the back-substitution that we were talking about earlier. Our u-sub did not get rid of all the ${\displaystyle x}$ terms in the integrand, so we need to back-sub to get rid of it. We find that ${\displaystyle x={\frac {u-3}{2}}.}$ After simplifying, we get the following. ${\displaystyle {\frac {1}{2}}\int _{3}^{5}x{\sqrt {u}}{\mathrm {d} }u={\frac {1}{4}}\int _{3}^{5}(u-3){\sqrt {u}}{\mathrm {d} }u}$ Expand and evaluate. An advantage when dealing with definite integrals is that you do not need to rewrite the antiderivative in terms of the original variable before evaluating. Doing so would introduce needless complications. ${\displaystyle {\begin{aligned}{\frac {1}{4}}\int _{3}^{5}(u-3){\sqrt {u}}{\mathrm {d} }u&={\frac {1}{4}}\int _{3}^{5}(u^{3/2}-3u^{1/2}){\mathrm {d} }u\\&={\frac {1}{4}}\left({\frac {2}{5}}u^{5/2}-3\cdot {\frac {2}{3}}u^{3/2}\right){\Bigg |}_{3}^{5}\\&={\frac {1}{4}}\left({\frac {2}{5}}(5)^{5/2}-2\cdot (5)^{3/2}-{\frac {2}{5}}(3)^{5/2}+2\cdot (3)^{3/2}\right)\\&={\frac {1}{4}}\left(-{\frac {6}{5}}\cdot 3^{3/2}+2\cdot 3^{3/2}\right)\\&={\frac {1}{4}}{\frac {4}{5}}3^{3/2}\\&={\frac {3{\sqrt {3}}}{5}}\end{aligned}}}$ EditIntegration by Parts See the main article on how to integrate by parts. The integration by parts formula is given below. The main goal of integration by parts is to integrate the product of two functions - hence, it is the analogue of the product rule for derivatives. This technique simplifies the integral into one that is hopefully easier to evaluate. ${\displaystyle \int u{\mathrm {d} }v=uv-\int v{\mathrm {d} }u}$ Evaluate the integral of the logarithm function. We know that the derivative of ${\displaystyle \ln x}$ is ${\displaystyle {\frac {1}{x}},}$ but not the antiderivative. It turns out that this integral is a simple application of integration by parts. ${\displaystyle \int \ln x{\mathrm {d} }x}$ Choose a ${\displaystyle u}$ and ${\displaystyle {\mathrm {d} }v,}$ and find ${\displaystyle {\mathrm {d} }u}$ and ${\displaystyle v}$. We choose ${\displaystyle u=\ln x}$ because the derivative is algebraic and therefore easier to manipulate. Then ${\displaystyle {\mathrm {d} }v={\mathrm {d} }x.}$ Therefore, ${\displaystyle {\mathrm {d} }u={\frac {1}{x}}{\mathrm {d} }x}$ and ${\displaystyle v=x.}$ Substituting all of these into the formula, we obtain the following. ${\displaystyle \int \ln x{\mathrm {d} }x=x\ln x-\int {\mathrm {d} }x}$ We converted the integral of a logarithm into the integral of 1, which is trivial to evaluate. Evaluate. ${\displaystyle \int \ln x{\mathrm {d} }x=x\ln x-x+C}$ EditVideo EditTips When defining the Riemann sum, the meaning of ${\displaystyle f(x_{i})}$ can mean the left-hand, right-hand, or midpoint of ${\displaystyle f(x_{i})}$ on the interval ${\displaystyle [x_{i},x_{i}+\Delta x].}$ These different definitions will give slightly different sums of the area of the rectangles. However, when the number of rectangles is sent to infinity, the error between any of these definitions goes to 0 and all the sums converge to the integral. This is what the diagram in step 2 actually shows - the blue rectangles are right-handed rectangles, the yellow are left-handed, the red sample the minimum on the interval, and the blue sample the maximum on the interval. The graph in the middle shows that all of these areas, when the number of rectangles is sent to infinity, converge to the integral. EditRelated wikiHows Integrate by Substitution Integrate by Parts Integrate by Partial Fractions Integrate by Differentiating Under the Integral Integrate Using the Gamma Function Integrate Gaussian Functions Integrate in Cylindrical Coordinates Integrate in Spherical Coordinates Calculate Contour Integrals Integrate Using Residue Theory

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