WEBVTT
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Antiderivatives
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At this point, you know how the calculate derivatives of many functions.
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And have been introduced to a variety of their applications.
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In this video, we want to ask a question that turns that process around.
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Given a function π, how do we find a function with the derivative π?
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And why would we be interested in such a function?
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The answer to the first part of that question is the antiderivative.
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That question being, given a function π, how do we find a function with the derivative π?
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The antiderivative of a function π is a function with the derivative π.
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In other words, itβs a function that reverses what the derivative does.
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Letβs consider the function π of π₯ equals two π₯.
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To take the antiderivative of this function, we need a function whose derivative is two π₯.
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We know that when we take the derivative of π₯ squared, we get two π₯.
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And that means the antiderivative of two π₯ would be π₯ squared.
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However, π₯ squared is not the only antiderivative of two π₯.
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π₯ squared plus one is also an antiderivative of two π₯.
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And thatβs because the derivative of a constant is zero.
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To account for this, we say that the antiderivative of two π₯ is π₯ squared plus π, where π is any constant value.
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Now letβs look at why we might be interested in doing something like this.
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Letβs consider what we know about motion.
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If we start with the position of an object and the function π of π‘, we can find the velocity of that object, the π£ of π‘, by taking the derivative of our position function.
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And if we take the derivative of our velocity function, we can calculate the acceleration of our object, the π of π‘.
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But what if we wanted to move in the other direction?
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If we knew our acceleration and we wanted to calculate the position, then we would need the antiderivative.
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This is just one of the many cases when we have a need for an antiderivative.
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Letβs move on and look at some examples of finding the antiderivative.
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Find the most general antiderivative capital πΉ of π₯ of the function lower case π of π₯ equals two π₯ to the seventh power minus three π₯ to the fifth power minus π₯ squared.
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To do this, weβll take the antiderivative of each of these terms separately.
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We need the antiderivative of two times π₯ to the seventh power.
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We need to know what function, when we take the derivative, equals two times π₯ to the seventh power.
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And for π₯ to the π power, we can find the antiderivative by taking π₯ to the π plus one power over π plus one.
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And then, in the general form, we add π to represent any constant.
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Letβs apply this to two times π₯ to the seventh power.
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Weβll leave two to the side, and weβll take π₯ to the seven plus one power and divide by seven plus one.
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The antiderivative of two times π₯ to the seventh power is two times π₯ to the eighth power over eight.
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And we can reduce this to one over four.
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Two times π₯ to the seventh power has an antiderivative of π₯ to the eighth power over four.
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And weβll repeat this process with negative three π₯ to the fifth.
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We can keep the negative three and we have π₯ to the five plus one power all over five plus one.
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Negative three times π₯ to the sixth power over six, which will reduce to π₯ to the sixth power over two.
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And weβll make sure that we keep that negative.
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Weβll repeat the process one final time.
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Weβre dealing with negative π₯ squared, so Iβll pull out a negative one.
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Then weβll have negative one times π₯ to the two plus one power over two plus one, negative π₯ cubed over three.
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Because weβre looking for the most general form, we canβt forget this constant at the end.
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Which makes our antiderivative πΉ of π₯ equals π₯ to the eighth power over four minus π₯ to the sixth power over two minus π₯ cubed over three plus π.
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Now letβs look at a case where we donβt want the most general form.
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Determine the antiderivative capital πΉ of the function lower case π of π₯ equals five π₯ to the fourth plus four π₯ cubed where capital πΉ of one equals negative two.
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Before we do anything else, weβll calculate the general antiderivative.
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And that means weβll follow the same process from the previous example.
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Weβll pull out the constant, add one to our exponent, and then divide by the value of the new exponent.
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In this case, weβll have five times π₯ to the fifth power divided by five.
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And weβll reduce that to π₯ to the fifth.
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Now, for the second term, take out that four, weβll raise π₯ cubed to π₯ to the fourth power, and then divide by four.
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Which reduces to π₯ to the fourth power.
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The four in the numerator and the denominator cancel out.
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If we were finding the general form, we would add a constant π.
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And we say that capital πΉ of π₯ equals π₯ to the fifth power plus π₯ to the fourth power plus π.
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And we wanna plug in πΉ of one to help us find the value of π.
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πΉ of one equals negative two.
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One to the fifth power plus one to the fourth power.
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One plus one equals two.
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So two plus π has to equal negative two.
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Subtract two from both sides.
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And we see that the constant value is negative four.
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Weβll take that information and plug it in to what we found for the general antiderivative.
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An antiderivative under these conditions is π₯ to the fifth plus π₯ to the fourth minus four.
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If the second derivative of π of π₯ equals three π₯ to the fifth plus three π₯ cubed plus five π₯ squared plus two, determine π of π₯.
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If weβre given a second derivative, we can take the antiderivative which will give us the first derivative of π of π₯.
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And then weβll take the antiderivative of that value which will give us our original π of π₯.
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The process wonβt be any different than our previous examples.
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Weβll just have to do that twice.
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Letβs take the antiderivative of three π₯ to the fifth which would become three times π₯ to the sixth power over six.
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And the second term, three times π₯ cubed, has an antiderivative of three times π₯ to the fourth over four.
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Five π₯ squared becomes five π₯ cubed over three.
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And the antiderivative of two is two π₯.
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Make sure you add your constant term.
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Now, at this point, we have the first derivative of this function.
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And we could simplify some of the coefficients here.
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But because weβre going to take the antiderivative again, Iβll wait and simplify in the last step.
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Now, we need the antiderivative of three times π₯ to the sixth power over six.
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Three-sixths times π₯ to the seventh power over seven.
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Three-fourths π₯ to the fourth is three-fourths times π₯ to the fifth over five.
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Five-thirds times π₯ cubed.
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Five-thirds times π₯ to the fourth over four.
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Two π₯ becomes two times π₯ squared over two.
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The antiderivative of a constant is that constant times π₯.
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And weβll need an additional constant on the end, which we can call π·.
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Weβve found our π of π₯ value, but we want to simplify.
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We can reduce this three over six to one-half.
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And then, weβll have π₯ to the seventh power over 14.
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We multiplied the denominators in our second term.
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And we get three times π₯ to the fifth over 20.
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Our third term is five times π₯ to the fourth over 12.
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In our fourth term, the twos cancel out and we have π₯ squared.
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Our final two terms, the ππ₯ plus π· canβt be simplified any further.
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Which means the general antiderivative of the function we were given is π₯ to the seventh power over 14 plus three times π₯ to the fifth power over 20 plus five times π₯ to the fourth power over 12 plus π₯ squared plus ππ₯ plus π·.
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For our last two examples, weβre going to consider what at first might seem like an irregular form.
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Determine the most general antiderivative capital πΉ of π₯ of the function lower case π, given that lower case π of π₯ equals five over two plus four over π₯.
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This first term, five over two, acts just like a constant.
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And we take its antiderivative by saying five over two times π₯.
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At first, it might not seem very clear what we can do with four over π₯.
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But what if we wrote it like this, four times of one over π₯?
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Now weβre saying, what function has a derivative of one over π₯?
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The natural log of π₯ has a derivative of one over π₯.
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This means the antiderivative of four times one over π₯ is going to be four times the natural log of π₯.
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Because four times the natural log of π₯ has a derivative of four times one over π₯.
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So weβll bring down four times the natural log of π₯.
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And since weβre considering a general form, weβll still need to add a constant value π.
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Capital πΉ of π₯ equals five over two π₯ plus four times the natural log of π₯ plus π.
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By considering the product rule, find the function π so that π prime of π₯ equals π to the π₯ power over the square root of π₯ plus two times π to the π₯ power times the square root of π₯.
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Weβll first need to remember the product rule for derivatives.
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That product rule tells us the derivative of the function π of π₯ times the function π of π₯ equals π of π₯ times the derivative of π of π₯ plus π of π₯ times the derivative of π of π₯.
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Before we try to find an π of π₯ and a π of π₯, letβs rewrite this function.
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We have π prime of π₯ equals π to the π₯ power.
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And we know that itβs being multiplied by one over the square root of π₯.
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We can write that as π₯ to the negative one-half power.
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Weβre multiplying π to the π₯ power times π₯ to the negative one-half power plus two times π to the π₯ power times π₯ to the one-half power.
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Something that we know is that the derivative of π to the π₯ power equals π to the π₯ power.
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If we say that π of π₯ equals π to the π₯ power, then π prime of π₯ also equals π to the π₯ power.
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This means that π₯ to the negative one-half power equals π prime of π₯.
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And it means that π of π₯ equals two times π₯ to the one-half power.
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π of π₯ equals two times π₯ to the one-half power.
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If we check that derivative, we get two times one-half times π₯ to the one-half minus one power, which is in fact π₯ to the negative one-half power.
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But what does this mean for us?
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Well, in the product rule, this value is the derivative of π of π₯ times π of π₯.
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And that means the antiderivative is going to be π of π₯ times π of π₯.
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We know π of π₯ and we know π of π₯, which means the antiderivative equals two times π₯ to the one-half power times π to the π₯ power.
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And we can put that back in the form it was given to us in.
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Two times the square root of π₯ times π to the π₯ power.
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Letβs briefly recap this key point about antiderivatives.
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If youβre given some function π prime of π₯, its antiderivative will be the function π of π₯ which derivative would then be π prime of π₯.
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Our very first example was when π prime of π₯ equals two π₯, we can take the antiderivative which gives us π₯ squared plus any constant π.
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And π₯ squared plus any constant π, if you take that derivative, would then give you two π₯.