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Find the most general antiderivative capital πΉ of π₯ of the function lowercase π of π₯ is equal to three π₯ squared minus two π₯ minus one.
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The question gives us the quadratic function lowercase π of π₯, and it wants us to find the most general antiderivative of this function lowercase π of π₯.
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Weβre told to call this most general antiderivative capital πΉ of π₯.
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Letβs start by recalling what an antiderivative is.
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We say that capital πΉ of π₯ is an antiderivative of lowercase π of π₯ if the derivative of capital πΉ of π₯ with respect to π₯ is equal to lowercase π of π₯.
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But we know antiderivatives of a function are not unique.
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For example, the derivative of π₯ plus one with respect to π₯ is equal to one, so π₯ plus one is an antiderivative of one.
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However, if we were also to calculate the derivative of π₯ minus one with respect to π₯, we would also get one.
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So π₯ minus one is also an antiderivative of one.
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In fact, for any constant π, the derivative of π₯ plus π with respect to π₯ will be one.
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We call this the most general antiderivative because we can input any value of π and get an antiderivative.
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So now we know the form our most general antiderivative will take.
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Letβs now discuss how weβll find the most general antiderivative of our quadratic function π of π₯.
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Letβs start by thinking about each time individually.
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Letβs start with three π₯ squared.
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We need the derivative of something to be equal to three π₯ squared.
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Recall the power rule for differentiation tells us for any constants π and π, the derivative of ππ₯ to the πth power with respect to π₯ is equal to π times π times π₯ to the power of π minus one.
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In other words, ππ₯ to the πth power is an antiderivative of πππ₯ to the power of π minus one.
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In our case, we want something which differentiates to give us an exponent of π₯ equal to two.
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So, weβll want our value of π equal to three.
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Letβs use π equal to three to try and find our value of π.
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Letβs differentiate ππ₯ cubed with respect to π₯.
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To do this, we use the power rule for differentiation.
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We multiply by our exponent of π₯ and then reduce this exponent by one.
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This gives us three ππ₯ squared.
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Remember, we want this to be an antiderivative of three π₯ squared.
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So we need our coefficient of π₯ squared equal to three.
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Therefore, weβll choose our value of π equal to one.
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Therefore, what weβve shown so far is π₯ cubed is an antiderivative of three π₯ squared.
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And we could in fact check this by differentiating π₯ cubed with respect to π₯.
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If we did this by using the power rule for differentiation, we would indeed get three π₯ squared.
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Letβs now do the same thing for our second term, negative two π₯.
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Again, we can try doing something very similar with the power rule for differentiation.
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This time, we need to notice that negative two π₯ is the same as negative two π₯ to the first power.
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So we want a function which differentiates to give us negative two π₯ to the first power.
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Therefore, we should set our value of π equal to two.
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So weβll set our value of π equal to two and try using the power rule for differentiation to differentiate ππ₯ squared with respect to π₯.
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Again, to do this, we multiply by our exponent of π₯ and then reduce this exponent by one.
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This gives us two ππ₯ to the first power.
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And of course, π₯ to the first power is just equal to π₯.
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Remember, weβre looking for an antiderivative of negative two π₯, so we need our coefficient of π₯ to be equal to negative two.
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And we see that this is indeed the case if our value of π is equal the negative one.
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What weβve shown is negative π₯ squared is an antiderivative of negative two π₯.
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And once again, we could verify this by differentiating negative π₯ squared with respect to π₯.
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We would use the power rule for differentiation, and we would indeed get negative two π₯.
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We could do exactly the same thing for our last term of negative one.
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But we already know the derivative of any linear function will just be the coefficient of π₯.
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So, just like we did in our earlier example, we can find an antiderivative of negative one by just having a linear function with the coefficient of π₯ equal to negative one.
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For example, negative π₯ will be an antiderivative of negative one.
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Once again, we could verify this by differentiating negative π₯ with respect to π₯.
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This is just the coefficient of π₯, which is negative one.
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Now that we found an antiderivative for each of the three terms in our function lowercase π of π₯, we can add these together to find an antiderivative of our function lowercase π of π₯.
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Remember, the reason this is true is if we were to differentiate this function, we could in fact evaluate the derivative term by term.
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And each of these terms is just the antiderivative of one of the terms of our function lowercase π of π₯.
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Therefore, this derivative will evaluate to give us lowercase π of π₯.
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But remember, the question is asking us to find the most general antiderivative.
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And to do this, we need to add a constant to our function.
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Weβll represent this with a capital πΆ.
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Therefore, we were able to show that the function lowercase π of π₯ is equal to three π₯ squared minus two π₯ minus one will have the most general antiderivative capital πΉ of π₯ is equal to π₯ cubed minus π₯ squared minus π₯ plus a constant πΆ.