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In this video, weβre going to learn how to find indefinite integrals of exponential and reciprocal functions.
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Weβll begin by recalling the first part of the fundamental theorem of calculus before looking at how this helps us to integrate exponential functions of the form π to the power of π₯ and reciprocal functions of the form π over π₯.
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We begin by stating the first part of the fundamental theorem of calculus.
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We let π be a continuous real-valued function defined on some closed interval π to π.
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We then let capital πΉ be the function defined for all π₯ in this closed interval by capital πΉ of π₯ equals the integral evaluated between π and π₯ of π of π‘ with respect to π‘.
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Then, capital πΉ is uniformly continuous on the closed interval π to π and differentiable on the open interval π to π such that capital πΉ prime of π₯ is equal to π of π₯ for all π₯ in the open interval π to π.
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In other words, capital πΉ is the antiderivative of a function π, the function whose derivative is equal to the original function.
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And essentially all this tells us is that integration is the reverse process to differentiation.
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So, letβs begin by looking at the function π of π₯ equals π to the power of π₯.
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Well, we know that the derivative of π to the power of π₯ is simply π to the power of π₯.
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So, the antiderivative of π to the power of π₯ is also π to the power of π₯.
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And we can, therefore, say that the indefinite integral of π to the power of π₯ evaluated with respect to π₯ two is π to the power of π₯.
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And of course, since weβre working with an indefinite integral, we add that constant of integration.
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Letβs call that π.
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But what about the integration of a function π to the power of π₯ for a real constant π?
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Well, once again, we think about derivatives.
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And we recall the fact that the derivative of π to the power of π₯ is π to the power of π₯ times the natural log of π.
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We can therefore say that the indefinite integral of π to the power of π₯ times the natural log of π is equal to π to the power of π₯, since thatβs its antiderivative.
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Once again, weβre working with an indefinite integral, so we add a constant of integration π.
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But this isnβt quite what weβre after.
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We were wanting to evaluate the integral of π of the power of π₯.
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So, we take this constant, the natural log of π, outside of our integral.
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And then, we divide both sides by the natural log of π.
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And weβve obtained the integral of π to the power of π₯ to be π to the power of π₯ over the natural log of π plus πΆ.
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Notice, too, that Iβve written this constant as capital πΆ.
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Thatβs to demonstrate that weβve divided our original constant by another constant, thereby changing this number.
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Now, itβs useful to know where these results come from, but generally itβs fine to state them.
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Weβre now going to look at some examples demonstrating the integration of these exponential functions.
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Determine the indefinite integral of eight π to the three π₯ minus π to the two π₯ plus nine over seven π to the power of π₯ with respect to π₯.
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Now, this question might look really nasty, and you might be starting to consider how a substitution might help.
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However, itβs important to notice that we can simplify the integrand by simply dividing each part of the numerator by seven π to the power of π₯, remembering that we can then simply subtract the exponents.
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When we do, weβre left with the indefinite integral of eight-sevenths π to the two π₯ minus one-seventh π to the π₯ plus nine-sevenths π to the negative π₯.
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Next, we recall that we can separate this integral.
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The integral of the sum of a number of functions is equal to the sum of the integrals of each function.
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And we can also take any constant factors outside of the integral and deal with them later.
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So, weβre looking at eight-sevenths of the integral of π to the two π₯ with respect to π₯ minus one-seventh of the integral of π to the π₯ with respect to π₯ plus nine-sevenths of the integral of π to the power of negative π₯ with respect to π₯.
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Okay, So what next?
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Well, we know that the indefinite integral of π to the power of π₯ is π to the power of π₯ plus π.
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But what about the integral of π of the power of two π₯?
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You might be wanting to make a prediction as to what you think this will give.
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And there is a standard result that we can quote.
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But letβs look at this derivation using integration by substitution.
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Weβre going to let π’ be equal to two π₯.
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So, dπ’ by dπ₯ is equal to two.
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Now, we know that 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 a half dπ’ is equal to dπ₯.
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And then, we see that we can simply replace two π₯ with π’ and dπ₯ with a half dπ’ and then take out this factor of one-half.
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And all we need to do now is integrate π to the power of π’ with respect to π’.
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Well, thatβs simply a half times π to the power of π’.
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But of course, since π’ is equal to two π₯, we can say that the indefinite integral of π to the power of two π₯ is a half times π to the power of two π₯ plus a constant of integration.
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Letβs call that π΄.
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And this is great because it provides us with the general result for the integral of π the power of ππ₯ for real constants π.
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Itβs one over π times π to the power of ππ₯.
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We can use this result.
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And we see that the integral of π to the power of negative π₯ is one over negative one times π to the power of negative π₯ plus π.
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So, popping this altogether, we see that our indefinite integral is eight-sevenths times a half π to the two π₯ plus π΄ minus a seventh times π to the π₯ plus π΅ plus nine-sevenths times one over negative one times π to the power of negative π₯ plus πΆ.
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Distributing our parentheses and combining all our constants, and weβre left with the solution as being four-sevenths π to the power of two π₯ minus π to power of π₯ over seven minus nine-sevenths π to the power of negative π₯ plus π·.
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This example demonstrated really nicely that whilst we can obtain the integral of π to the power of ππ₯ from the general result for π of the power of π₯ and a clever substitution.
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Itβs much more sensible to use the result given.
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That is the integral of π to the power of ππ₯ with respect to π₯ is one over π times π to the power of ππ₯ plus π.
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We can even generalize further and say that the integral of π to the power of ππ₯ plus π is π the power of ππ₯ plus π over π.
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In our next example, weβre going to consider how a substitution can help us obtain the result for the integral of π to the power of ππ₯, where both π and π are real constants.
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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.
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This example gives us a general result for the integration of π to the power of ππ₯ for real constants π and π.
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Itβs π to the power of ππ₯ over π times the natural log of π plus that constant of integration πΆ.
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Weβre now going to consider some reciprocals.
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That is, functions of the form π over π₯.
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Now, you might initially think that this is quite straightforward.
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We can rewrite this function as π times π₯ to the power of negative one and go from there.
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But letβs see what happens when we use the power rule for integration.
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We add one to the power.
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That gives us π₯ to the power of zero.
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And then, we divide by zero.
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We canβt do that though.
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It gives us a number which is undefined.
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So, what else can we do?
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Well, once again, weβre going to consider a derivative.
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We recall the derivative of the natural log of π₯ as being one over π₯.
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And this leads us really nicely into the result that the integral of the one over π₯ is the natural log of π₯.
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But actually, we do need to redefine this a little.
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The natural log of π₯ can only take real positive values of π₯ is greater than zero.
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So, we can instead say that the integral of one over π₯ for non-zero values of π₯ is the natural log of the absolute value of π₯.
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And of course, we have that constant of integration π.
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Generalizing for π over π₯, we find that the integral of π over π₯ is π times the natural log of the absolute value of π₯ plus π, of course.
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Determine the indefinite integral of negative two over seven π₯ dπ₯.
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Weβre going to begin by removing the constant factor from this expression.
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And that gives us negative two-sevenths times the integral of one over π₯ dπ₯.
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We then quote the general result for the integral of one over π₯.
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Itβs the natural log of the absolute value of π₯.
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And so, we obtain our integral to be negative two-sevenths times the natural log of the absolute value of π₯ plus π.
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Finally, we distribute our parentheses.
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And we find that the solution to this question is negative two-sevenths times the natural log of the absolute value of π₯ plus capital πΆ.
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And notice here Iβve written capital πΆ to demonstrate that the original constant has been multiplied by negative two-sevenths, thereby changing it.
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Itβs useful to remember that we can actually check our solution by performing the reverse process, by differentiating.
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The derivative of the natural log of π₯ is, of course, one over π₯.
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So, the derivative of negative two-sevenths times the natural log of π₯ is negative two-sevenths times one over π₯.
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And the derivative of a constant πΆ is zero.
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Multiplying, and we end up with our derivative to be negative two over seven π₯, as required.
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Find, if possible, an antiderivative capital πΉ of π of π₯ equals one over two π₯ minus one that satisfies the conditions capital πΉ of zero is one and capital πΉ of one is negative one.
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Another way of thinking about the antiderivative is evaluating the indefinite integral.
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So, what weβre actually going to do is integrate one over two π₯ minus one with respect to π₯ and consider the conditions on it in a moment.
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To integrate one over two π₯ minus one, weβre going to use a substitution.
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Weβre going to let π’ be equal to two π₯ minus one, which means that dπ’ by dπ₯ equals two.
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Now, we do know that dπ’ by dπ₯ is not a fraction, but we do treat it a little like one for the purposes of integration by substitution and say that a half dπ’ equals dπ₯.
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Letβs perform some substitutions.
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We replace two π₯ minus one with π’ and dπ₯ with a half dπ’.
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And we can, of course, take out the constant factor of one-half, and we have half times the integral of one over π’.
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Well, remember, the integral of one over π₯ with respect to π₯ is the natural log of the absolute value of π₯.
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So, we can say that the integral of one over two π₯ minus one dπ₯ is a half times the natural log of π’ plus π.
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We replace π’ with two π₯ minus one, and weβve obtained our antiderivative.
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Distributing the parentheses, we see that the antiderivative π is equal to a half times the natural log of two π₯ minus one plus capital πΆ.
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Weβre now going to consider conditions.
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But we also recall the fact that two π₯ minus one cannot be equal to zero.
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So, weβre actually going to have a piecewise function for our antiderivative.
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The first part of the function weβre interested in is when two π₯ minus one is greater than zero.
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Solving for π₯, and we obtain π₯ to be greater than one-half.
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The second is when one minus two π₯ is greater than zero.
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Solving for π₯ this time, and we obtain π₯ to be less than one-half.
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Letβs clear some space and formalize this.
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We currently have that our antiderivative, capital πΉ of π₯, is equal to a half times the natural log of one minus two π₯ plus some constant when π₯ is less than one-half, and a half times the natural log of two π₯ minus one plus some constant when π₯ is greater than one half.
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Weβre now going to use our conditions on the antiderivative to evaluate π.
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The first is capital πΉ of zero equals one.
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So, when π₯ is equal to zero, πΉ is equal to one.
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Since π₯ is less than one-half, weβre going to use the first part of our function.
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We substitute π₯ equals zero and capital πΉ equals one in.
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And we get one equals a half times the natural log of one minus two times zero.
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Well, the natural log of one minus two times zero is the natural log of one, which is of course zero.
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So, we obtain π here to be one.
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The second condition we have is that when π₯ is equal to one, capital πΉ equals negative one.
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This time π₯ is greater than one-half, so weβre going to use the second part of the function.
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We see that negative one is equal to a half times the natural log of two times one minus one plus π.
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Once again, the natural log of two times one minus one is zero.
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So, we see that π in the second part is equal to negative one.
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And so, the antiderivative capital πΉ does indeed exist.
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Such that capital πΉ of π₯ is equal to a half times the natural log of one minus two π₯ plus one for π₯ is less than one-half and a half times the natural log of two π₯ minus one minus one for π₯ is greater than one-half.
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In this video, weβve seen that we can use the first part of the fundamental theorem of calculus to integrate exponential reciprocal functions.
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We saw that the integral of π the power of ππ₯ is equal to one over π times π to the power of ππ₯.
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The integral of π to the power of ππ₯ is π to the power of the ππ₯ over π times the natural log of π.
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And the integral of π over π₯ with respect to π₯ is π times the natural log of the absolute value of π₯, given that π₯ is not equal to zero.
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And of course, since weβve been working with indefinite integrals, we need to assume that weβve got this constant of integration π each time.