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Given that π double prime of π₯ is equal to negative five π to the power of four π₯ plus two π₯ to the fifth power, find π of π₯.
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Recall, first of all, that π double prime of π₯ means the second derivative of π with respect to π₯.
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So the function weβve been given is the result of differentiating π twice.
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In order to find the function π of π₯, we need to go back the other way.
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And as integration is the reverse of differentiation, we therefore need to integrate this function twice.
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Integrating once will give us the first derivative of our function π of π₯.
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So we have that π prime of π₯ is equal to the integral of negative five π to the power of four π₯ plus two π₯ to the fifth power with respect to π₯.
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Now, there are two types of functions that we need to recall how to integrate in order to answer this question.
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The more straightforward of these is just a general polynomial term.
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And so, we recall that the integral or antiderivative of π₯ to the power of π, where π is some real number not equal to negative one, is π₯ to the power of π plus one over π plus one plus a constant of integration πΆ.
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We increase the exponent by one and then divide by the new exponent.
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The second type of function we need to recall how to integrate is an exponential function or, more specifically, a function of the form π to the power of ππ₯ for some constant π.
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We recall that this integral is equal to one over π π to the power of ππ₯ plus a constant of integration πΆ.
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The factors of negative five and positive two just act as multiplicative constants.
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So letβs perform this integral.
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We have negative five, and then we integrate π to the power of four π₯, which weβve seen is equal to one-quarter π to the power of four π₯.
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We then have plus two, and we integrate π₯ to the power of five, which weβve just seen is π₯ to the power of six over six.
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Weβll only include one single constant of integration which weβll call πΆ.
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Simplifying the constants, and we have that π prime of π₯ is equal to negative five over four π to the power of four π₯ plus one-third, that simplified from two-sixths, π₯ to the sixth power plus some constant πΆ.
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Remember, though, that weβre looking to find the function π of π₯.
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And, so far, we have π prime of π₯.
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So we need to integrate a second time.
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We have then that π of π₯ is equal to the integral of negative five over four π to the power of four π₯ plus one-third π₯ to the power of six plus πΆ with respect to π₯.
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And we can now apply the two rules of integration that weβve already written down for a second time.
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Integrating the first two terms, we have negative five over four multiplied by a quarter π to the power of four π₯ plus one-third multiplied by one-seventh π₯ to the seventh power.
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We then need to integrate that constant πΆ, and this will give us πΆπ₯.
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We can think of πΆ as πΆπ₯ to the power of zero.
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So when we increase the exponent, we get π₯ to the power of one.
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And then we divide by one.
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We must also remember to include a new constant of integration which weβll call π·.
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Simplifying each of the constants, and we have our answer to the problem.
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π of π₯ is equal to negative five over 16 π to the power of four π₯ plus one over 21 π₯ to the seventh power plus πΆπ₯ plus π·, where πΆ and π· are constants.