WEBVTT
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Indefinite Integrals: The Power Rule
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In this video, we will learn how to find the indefinite integrals of polynomials and general power functions using the power rule for integration.
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Letβs start by recalling what the antiderivative of a function is.
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We can say that capital πΉ is the antiderivative of lowercase π if capital πΉ prime of π₯ is equal to lowercase π of π₯.
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We can in fact say that this is true for any function π of π₯ where π of π₯ is equal to capital πΉ of π₯ plus πΆ for any constant πΆ.
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Now, this is very useful as weβll be using it to define an indefinite integral.
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We can say that the indefinite integral of lowercase π of π₯ with respect to π₯ is equal to capital πΉ of π₯ plus πΆ where capital πΉ is the antiderivative of lowercase π.
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Itβs very important to remember our constant of integration when performing an indefinite integral.
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Letβs quickly consider why this constant is here.
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If we perform the reverse operation on this equation, so thatβs differentiating with respect to π₯, then since we are simply performing the inverse operation on the left, the differential with respect to π₯ of the integral of lowercase π of π₯ with respect to π₯ will just be lowercase π of π₯.
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On the right-hand side, when we differentiate capital πΉ of π₯ with respect to π₯, we get π prime of π₯ and our constant πΆ will simply vanish because differentiating a constant gives zero.
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Now, when we go back the other way to our integral, the constant will appear again.
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However, we do not know the value of this constant, hence why we label it πΆ.
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Itβs just an unknown constant.
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Letβs now consider what happens when we integrate some function with respect to π₯, for example, three π₯.
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We know the equation for finding an indefinite integral.
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We have that the integral of lowercase π of π₯ with respect to π₯ is equal to uppercase πΉ of π₯ plus πΆ where uppercase πΉ is the antiderivative of lowercase π.
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Now in our case, lowercase π of π₯ is equal to three π₯.
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So we simply need to find the antiderivative of three π₯.
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Letβs try and do this by trial and error.
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Weβre trying to work out what we need to differentiate in order to get three π₯.
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Letβs start by differentiating π₯ squared with respect to π₯.
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This gives us two π₯ which is very close to three π₯.
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However, weβre not quite there yet.
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Letβs try multiplying our differential by one-half.
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Differentiating π₯ squared over two with respect to π₯ give us π₯ which is one-third of what weβre trying to find.
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So letβs now try multiplying our differential by three.
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And differentiating three π₯ squared over two with respect to π₯ gives us three π₯.
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Therefore, we found our antiderivative.
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And itβs three π₯ squared over two.
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So this is all capital πΉ of π₯.
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Therefore, we can say that our integral is equal to three π₯ squared over two plus πΆ.
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Now this is quite a lengthy method for finding antiderivatives of power functions.
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Letβs consider the power rule for derivatives.
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We know that the derivative of π₯ to the power of π plus one with respect to π₯ is π plus one multiplied by π₯ to the power of π since for the power rule of derivatives we multiply by the power and then decrease the power by one.
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What this tells us is that π₯ to the power of π plus one is the antiderivative of π plus one multiplied by π₯ to the power of π.
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We can therefore recall π₯ to the power of π plus one capital πΉ of π₯ and π plus one multiplied by π₯ to the power of π lowercase π of π₯.
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And we can substitute these functions into our equation for the indefinite integral of a function.
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And this gives us that the integral of π plus one multiplied by π₯ to the power of π with respect to π₯ is equal to π₯ to the power of π plus one plus πΆ.
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Now, we have a constant inside our integral.
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And we can, in fact, factor this out.
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Letβs quickly consider why we can do this.
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We know that if we differentiate a constant multiplied by some function π of π₯, then this is equal to the constant multiplied by the differential of π of π₯ with respect to π₯.
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This, therefore, tells us that the integral of π multiplied by π of π₯ with respect to π₯ is equal to π multiplied by the integral of π of π₯ with respect to π₯.
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Therefore weβre able to factor out our constant of π plus one.
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And then we can simply divide by this constant.
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And this gives us that the integral of π₯ to the power of π with respect to π₯ is equal to π₯ to the power of π plus one over π plus one plus πΆ over π plus one.
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However, since both πΆ and π plus one are constants, this means that πΆ over π plus one will also be a constant.
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And we can call that constant π·.
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Therefore, we arrive at our power rule for integration.
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It tells us that the integral of π₯ to the power of π with respect to π₯ is equal to π₯ to the power of π plus one over π plus one plus πΆ.
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Here I have relabeled our constant of integration back to πΆ just for consistency with our definition of an indefinite integral.
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However, it does not matter what you label this constant.
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An easy way to remember this rule is that when we integrate a power function, we increase the power by one and then divide by the new power.
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And of course, we mustnβt forget to add our constant of integration.
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Now, this power rule for integration works for any real value of π except for one specific value.
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And that is when π is equal to negative one.
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If we tried using π is equal to negative one, we get that the indefinite integral of π₯ to the power of negative one with respect to π₯ is equal to π₯ to the power of negative one plus one over negative one plus one plus πΆ.
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Since negative one plus one is zero, this means that we have a zero in the denominator of our fraction.
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Therefore, this fraction is undefined and so is the integral.
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And so we can say that our power rule for integration works for all real values of π except negative one.
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Now it is, in fact, possible to integrate π₯ to the power of negative one with respect to π₯.
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However, we will not be covering it in this video.
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We will now be moving on to an example of how we can use the power rule.
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Determine the indefinite integral of negative π₯ to the power of nine with respect to π₯.
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In order to find this integral, we need to use the power rule for integration.
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We have that the integral of π₯ to the power of π with respect to π₯ is equal to π₯ to the power of π plus one over π plus one plus πΆ.
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And weβre trying to find the integral of negative π₯ to the power of nine with respect to π₯.
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Letβs start by factoring out the negative one.
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We have that our integral is equal to the negative integral of π₯ to the power of nine with respect to π₯.
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Now, when we look at our integral and we compare it to the integral in the formula, we can see that in our case π is equal to nine.
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Therefore, we simply substitute π is equal to nine into the formula.
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Therefore, we obtain negative π₯ to the power of nine plus one all over nine plus one plus πΆ.
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This gives negative π₯ to the power of 10 over 10 minus πΆ.
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Now, negative πΆ is just another constant.
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So we can call it π·.
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And now weβve reached our solution.
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And that is that the indefinite integral of negative π₯ to the power of nine with respect to π₯ is equal to negative π₯ to the power of 10 over 10 plus π·.
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Letβs now work out how we will integrate a polynomial.
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We know that the differential of a sum of functions, so π of π₯ plus π of π₯ with respect to π₯, is equal to the sum of the differential of the functions.
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So π by dπ₯ of π of π₯ plus π by dπ₯ of π of π₯.
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From this we obtain that the integral of a sum of functions is equal to the sum of the integrals of the functions.
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Using this rule, weβre able to split integrals of polynomials down into integrals of power functions.
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For example, the integral of π₯ squared plus π₯ with respect to π₯ is equal to the integral of π₯ squared with respect to π₯ plus the integral of π₯ with respect to π₯.
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And we already know how to integrate these power functions on the right.
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Using this, weβre able to integrate any polynomial.
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Letβs now look at an example.
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Determine the indefinite integral of 25π₯ squared minus 65π₯ plus 36 with respect to π₯.
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Letβs start by splitting this integral up using the fact that the integral of a sum of functions is equal to the sum of the integrals of the functions.
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We obtain that our integral is equal to the integral of 25π₯ squared with respect to π₯ plus the integral of negative 65π₯ with respect to π₯ plus the integral of 36 with respect to π₯.
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We can factor out the constant in each integral.
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Now, we can use the power rule for integration which tells us that the integral of π₯ to the power of π with respect to π₯ is equal to π₯ to the power of π plus one over π plus one plus πΆ.
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In the case of our first integral, π is two.
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Therefore, itβs equal to 25 multiplied by π₯ cubed over three plus some constant which weβll call πΆ one.
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For our second integral, weβre integrating π₯.
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π₯ is equal to π₯ to the power of one.
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Therefore, π is equal to one.
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Therefore, itβs equal to negative 65 multiplied by π₯ squared over two plus some constant which weβll call πΆ two.
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For our third integral, weβre integrating one.
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One is also equal to π₯ to the power of zero.
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Therefore, our value of π is zero.
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And so itβs equal to 36 multiplied by π₯ to the power of one over one plus some constant which weβll call πΆ three.
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Expanding and simplifying, we obtain that our integral is equal to 25π₯ cubed over three minus 65π₯ squared over two plus 36π₯ plus 25πΆ one minus 65πΆ two plus 36πΆ three.
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Now, since πΆ one, πΆ two, and πΆ three are all constant, 25πΆ one minus 65πΆ two plus 36πΆ three is also a constant.
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And we can relabel this as πΆ.
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And so we reach our solution.
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And that is that the indefinite integral of 25π₯ squared minus 65π₯ plus 36 with respect to π₯ is equal to 25π₯ cubed over three minus 65π₯ squared over two plus 36π₯ plus πΆ.
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In the next example, weβll see how we can simply integrate a polynomial without splitting it up into separate integrals.
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Determine the indefinite integral of π₯ minus six multiplied by π₯ minus five multiplied by π₯ minus three.
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Letβs start by expanding the brackets.
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Expanding the first two sets of brackets, we get π₯ squared minus 11π₯ plus 30.
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And we then multiply this by π₯ minus three.
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We obtain the indefinite integral of π₯ cubed minus 14π₯ squared plus 63π₯ minus 90 which is in fact a polynomial.
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And we can use the power rule for integration to integrate this term by term.
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The power rule tells us that the integral of π₯ to the power of π with respect to π₯ is equal to π₯ to the power of π plus one over π plus one plus πΆ.
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If instead we were integrating π₯ to the power of π multiplied by some constant π, then since we can factor a constant out of our integral, then this would simply be equal to π multiplied by π₯ to the power of π plus one over π plus one plus πΆ.
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Now, you may be wondering why we havenβt multiplied the πΆ by π.
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And thatβs because π is also a constant.
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Therefore, π multiplied by πΆ is a constant.
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And we can just rename this new constant to be πΆ.
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So now, letβs apply this rule to our integral term by term.
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The first term is π₯ cubed.
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Therefore, π is equal to three.
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We increase the power by one and divide by the new power to get π₯ to the power of four over four.
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The next term is negative 14π₯ squared.
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Negative 14 is just a constant.
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So that will remain.
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Our power is two.
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So π is two.
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We increase the power by one to get π₯ cubed and divide by the new power.
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On next term, we have 63π₯.
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So we can start by writing in our constant of 63.
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We then note that π₯ is equal to π₯ to the power of one.
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So π is equal to one.
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So when we integrate this, we get π₯ squared over two.
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For our final term, we have negative 90.
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And we know that this can also be written as negative 90 multiplied by π₯ to the power of zero since π₯ to the power of zero is just one.
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So now, we can integrate it.
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We start by writing our constant of negative 90.
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Since our power of π₯ is zero, we increase the power by one giving us π₯ to the power of one and divide by the new power.
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So thatβs dividing by one.
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And we miss then adding our constant of integration πΆ.
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We can write this out a little neater for our solution which is π₯ to the power of four over four minus 14π₯ cubed over three plus 63π₯ squared over two minus 90π₯ plus πΆ.
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Letβs quickly note that this integration power rule works for any real π except negative one.
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So that includes both negative and noninteger powers of π₯.
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Letβs see how this works in the following examples.
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Determine the indefinite integral of negative two over seven multiplied by π₯ to the power of negative nine with respect to π₯.
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In our integral, we simply have a power function.
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So we can use the power rule for integration in order to find this integral.
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It tells us that the indefinite integral of π₯ to the power of π with respect to π₯ is equal to π₯ to the power of π plus one over π plus one plus πΆ.
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In our case, weβre integrating negative two-sevenths π₯ to the power of negative nine with respect to π₯.
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So our power of π₯ is negative nine.
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We can start by writing down our constant which is negative two-sevenths.
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Then since our value of π is negative nine, we next need to write π₯ to the power of π plus one over π plus one.
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So thatβs π₯ to the power of negative nine plus one which is π₯ to the power of negative eight over negative nine plus one.
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So thatβs negative eight.
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Then we mustnβt forget to add our constant of integration πΆ.
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For our final step here, we just need to simplify.
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And we obtain our solution which is that the indefinite integral of negative two-seventh multiplied by π₯ to the power of negative nine with respect to π₯ is equal to π₯ to the power of negative eight over 28 plus πΆ.
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In our final example, weβll see how to integrate a function with noninteger powers of π₯.
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Determine the indefinite integral having negative four multiplied by the fifth root of π₯ to the power of nine plus eight all multiplied by the fifth root of π₯ squared with respect to π₯.
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Letβs start by writing this roots as powered.
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We know that the πth root of π₯ is equal to π₯ to the power of one over π.
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Once weβve written our roots as powers, we can then combine them with the existing powers, using the fact that π₯ to the power of π to the power of π is equal to π₯ to the power of π multiplied by π.
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Therefore, π₯ to the power of nine to the power of one-fifth becomes π₯ to the power of nine-fifths.
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And π₯ squared to the power of one-fifths becomes π₯ to the power of two-fifths.
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Now, we can expand the brackets, using the fact that π₯ to the power of π times π₯ to the power of π is equal to π₯ to the power of π plus π.
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So our integral becomes the integral of negative four π₯ to the power of eleven-fifths plus eight multiplied by π₯ to the power of two-fifths with respect to π₯.
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Here, we can use the power rule for integration which tells us that the indefinite integral of π₯ to the power of π with respect to π₯ is equal to π₯ to the power of π plus one over π plus one plus πΆ.
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We can apply this rule to our integral term by term.
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For the first term, we have negative four π₯ to the power of eleven-fifths.
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Therefore, π is equal to eleven-fifths.
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When we integrate this term, we get negative four multiplied by π₯ to the power of π plus one.
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And π plus one is simply sixteen-fifths.
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So itβs π₯ to the power of sixteen-fifths.
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And then we need to divide by π plus one.
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So thatβs dividing by sixteen-fifths.
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For the second time, we have eight multiplied by π₯ to the power of two-fifths.
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Therefore, π is two-fifths.
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So we add eight multiplied by π₯ to the power of π plus one which is π₯ to the power of seven-fifths.
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And we then divide by seven-fifths.
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And we mustnβt forget to add our constant of integration πΆ.
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Now, all that remains to do is to simplify.
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And so we obtain a solution that the indefinite integral of negative four multiplied by the fifth root of π₯ to the power of nine plus eight all multiplied by the fifth root of π₯ squared with respect to π₯ is equal to negative five multiplied by π₯ to the power of sixteen-fifths over four plus 40 multiplied by π₯ to the power of seven-fifths over seven plus πΆ.
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Weβve now covered a variety of examples of indefinite integrals of power functions.
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Letβs recap some key points of the video.
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Key points.
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The indefinite integral of lowercase π of π₯ with respect to π₯ is equal to capital πΉ of π₯ plus πΆ where capital πΉ of π₯ is the antiderivative of lowercase π of π₯ and πΆ is a constant.
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The power rule for integration.
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The indefinite integral of π₯ to the power of π with respect to π₯ is equal to π₯ to the power of π plus one over π plus one plus πΆ for any real number π except negative one.