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Determine the integral of two to the power of nine π₯ with respect to π₯.
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Letβs begin by quoting what we do know about the integral of π to the power of π₯.
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Itβs π to the power of π₯ divided by the natural log of π.
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Our integrand is slightly different though; itβs a constant to the power of another constant times π₯.
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So, weβre going to use the process of introducing something new, a new letter.
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We let π’ be equal to nine π₯, and of course this is known as integration by substitution.
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We obtain the derivative of dπ’ with respect to π₯ to be equal to nine.
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Now, remember, dπ’ by dπ₯ is absolutely not a fraction, but we can treat it a little like one for the purposes of integration by substitution.
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And we see that we can say that a ninth dπ’ equals dπ₯.
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We replace π’ with nine π₯ and dπ₯ with a ninth dπ’.
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And then, we take out this constant factor of a ninth.
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And we see that our integral is now a ninth of the integral of two to the power of π’ with respect to π’.
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Well, the integral of two to the power of π’ is two to the power of π’ over the natural log of two.
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And then, of course, we can use the definition of our substitution and replace π’ with nine π₯.
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And weβve found the integral of two to the power of nine π₯ with respect to π₯.
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Itβs two to the power of nine π₯ over nine times the natural log of two plus this constant of integration πΆ, which Iβve made a capital πΆ to show that itβs different from the value we had before.