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The fundamental theorem of calculus: evaluating definite integrals.
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In this video, we will learn how to evaluate definite integrals using the fundamental theorem of calculus.
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The theorem is usually stated in two parts.
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For the purpose of this video, weβll be focusing on the second part.
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This tells us that if lowercase π is a continuous function on the closed interval between π and π and capital πΉ is any antiderivative of lowercase π.
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Here expressed as capital πΉ prime of π₯ is equal to lowercase π of π₯.
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Then the integral between π and π of lowercase π of π₯ with respect to π₯ is equal to uppercase πΉ of π minus uppercase πΉ of π.
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Now here, an interesting point to note is that weβve said capital πΉ is any antiderivative of lowercase π.
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This means that if there are many antiderivatives, it does not matter which one we pick to use our theorem.
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To make sense of this, let us think back to what we mean by an antiderivative.
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By now, we should be familiar with the fact that the first part of the fundamental theorem of calculus essentially tells us that differentiation and integration are inverse processes.
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This means that we can find the general form for the antiderivative of a function, lowercase π of π₯, by evaluating the indefinite integral of that function.
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Letβs consider an example function.
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Lowercase π one of π₯ is equal to two π₯.
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Its antiderivative, uppercase πΉ one of π₯, would be equal to the indefinite integral of two π₯ with respect to π₯.
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To solve this, we use the power rule of integration, raising the power of π₯ by one and dividing by the new power.
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Of course, here, we mustnβt forget to add on our constant of integration, π.
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More simply, our answer is written as π₯ squared plus π.
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Now, this constant of integration π can take any value we like.
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And our expression would still be one of an infinite number of antiderivatives of our original function πΉ one of π₯.
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If we took the case of π equals zero, πΉ one of π₯ would simply be π₯ squared.
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π equals five would give us an antiderivative of π₯ squared plus five.
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π could even be negative π, which would mean our antiderivative would be π₯ squared minus π.
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All of these are antiderivatives of two π₯.
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Given that the fundamental theorem of calculus allows us to use any of these antiderivatives to evaluate a definite integral, what do we do.
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Letβs continue to use our example function π one of π₯ equals two π₯.
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That now, weβre looking to the definite integral between π and π of this function.
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The theorem tells us that this is equal to the antiderivative capital πΉ one of π minus capital πΉ one of π.
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We can arbitrarily pick one of our antiderivatives, for example, π₯ squared plus five.
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If πΉ one of π₯ equals π₯ squared plus five, then πΉ one of π equals π squared plus five.
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Similar reasoning follows for πΉ one of π.
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Simplifying this, we see that there is a plus five and a minus five term, which cancel each other out.
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And weβre therefore left with π squared minus π squared.
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In fact, if we had used the general form which would involve any constant π, the same thing wouldβve happened.
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So regardless of the constant we use, we arrive at the same result.
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Given this fact, we can simply choose to ignore the constant of integration altogether when evaluating a definite integral.
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And this is essentially equivalent to taking the case where π is equal to zero.
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If we had considered this case to begin with, we wouldβve reached the same answer, but with fewer steps of working.
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Let us now see an example of our theorem in practice.
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Let π of π₯ equal six π₯ squared plus one.
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Evaluate the definite integral of π from π₯ equals two to π₯ equals three.
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For this question, weβve been asked to evaluate a definite integral, which using standard notation would look like this.
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We see that the integrand is our function π of π₯.
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And the limits of integration are two and three, as specified by the question.
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To evaluate this definite integral, weβre gonna be using the second part of the fundamental theorem of calculus.
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This tells us that if lowercase π is a continuous function on the closed interval between π and π and capital πΉ prime of π₯ is equal to lowercase π of π₯.
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In other words, capital πΉ is an antiderivative of lowercase π.
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Then the integral between π and π of lowercase π of π₯ with respect to π₯ is equal to capital πΉ of π minus capital πΉ of π.
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Letβs now apply this theorem to solve our problem.
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Our function, lowercase π of π₯, is equal to six π₯ squared plus one.
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To find this antiderivative, capital πΉ of π₯, we can integrate lowercase π of π₯.
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If we apply the power rule of integration, raising the power of π₯ in each of our terms and then dividing by the new power, we get an answer of six π₯ cubed over three plus π₯.
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And we also add our constant of integration π.
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At this point, we remember that the fundamental theorem of calculus allows us to use any antiderivative, which means π can take any value.
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It makes sense for us to choose the simplest case possible, which is when π is equal to zero.
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Essentially, this means we can ignore this constant.
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Simplifying, the antiderivative that weβll then use is two π₯ cubed plus π₯.
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Great.
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Letβs set this antiderivative to one side and go back to our original calculation.
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The question has asked us to evaluate this definite integral.
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Our upper limit of integration is three.
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And our lower limit is two.
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The fundamental theorem of calculus then tells us that this integral is equal to capital πΉ of three minus capital πΉ of two.
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Now, since we have just found capital πΉ of π₯, our antiderivative, we can substitute in the values of three and two into this function.
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After we have input these values, we can perform a few simplifications on our new expression.
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After working through these simplifications, we reach an answer of 39.
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With this step, weβve completed our question.
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The definite integral given in the question evaluates to 39.
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We evaluated our integral using the antiderivative of the given function πΉ of π₯.
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And the tool that we used to help us was the second part of the fundamental theorem of calculus.
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Okay, before moving on, a quick word on notation.
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Given the definite integral between π and π of a function π of π₯ with respect to π₯, very often, youβll see the next step written in the following form.
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With the antiderivative of the given function written in brackets like so and the limits of integration carried over to the right-hand bracket.
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And this is simply a shorthand.
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Itβs a way of expressing the antiderivative as a function before we substitute in our limits of integration and evaluate.
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It is equivalent to what we should now be familiar with, which is capital πΉ of π minus capital πΉ of π.
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Youβll almost always see this step, since itβs a very useful shorthand, which helps to tidy up our working.
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Looking back at our previous example function, π one of π₯ is equal to two π₯.
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If we were to consider the definite integral between one and three of this function with respect to π₯, our next step would look something like this, with the antiderivative of π one of π₯ in brackets, as shown.
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We would then proceed to continue our evaluation by substituting in three and one, the limits of integration.
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And weβd eventually reach an answer of eight.
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Let us look at an example question using this notation.
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Evaluate the integral between zero and two of two sin π₯ minus three π to the π₯ with respect to π₯.
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To answer this question, weβre going to be using the second part of the fundamental theorem of calculus.
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This tells us that if lowercase π is a continuous function on the closed interval between π and π and capital πΉ is any antiderivative of lowercase π.
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Then the integral between π and π of lowercase π of π₯ with respect to π₯ is equal to capital πΉ of π minus capital πΉ of π.
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Looking back at our question, the first thing we might notice is that the function lowercase π, which is our integrand, consists of two different terms.
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The first term involves the trigonometric function sine.
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And the second involves the exponential π.
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Now, we should be familiar with the fact that both trigonometric and exponential functions of this form are continuous over the entire set of real numbers.
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We have therefore fulfilled the criteria that our function π must be continuous over the closed interval between π and π, which in our case is the closed interval between zero and two.
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Now, given that we do have two terms, we might find that our working is clearer if we split these up into separate integrals.
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We would do this like so, remembering to keep the limits of integration the same across both terms.
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We can now evaluate each of these integrals separately.
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The antiderivative of two sin π₯ is negative two cos π₯.
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And the antiderivative of negative three π to the π₯ is negative three π to the π₯.
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Of course, remember, we can ignore the constant of integration in both cases since weβre working with definite integrals.
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Here, we note that weβve expressed our antiderivative in brackets, with the limits of integration being carried over to the right-hand bracket in both cases.
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Given that both of these brackets have the same limits of integration, we can simply combine them.
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Now, you might notice that we couldβve moved directly from our original integral to this step, by treating each of the terms individually.
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Instead of splitting our integral into two and then recombining, we would simply have found the antiderivative of each of our terms.
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If youβre not sure, however, thereβs no harm in writing out the method in full.
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To move forward with our question, we now substitute in the limits of our integration, which are zero and two.
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We then reach the following expression.
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With our first set of parentheses, thereβre no simplifications necessary.
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So we can just leave this.
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For the second set of parentheses, we might recall that cos of zero is equal to one.
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So negative two cos zero is equal to negative two.
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Alongside this, π to the power of zero is also one.
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So negative three π to the power of zero is negative three.
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Our second set of parentheses therefore becomes negative two minus three, which is negative five.
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But for our final answer, weβre subtracting this.
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So weβre left with a positive five.
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And weβve now reached our final answer.
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The definite integral given in the question evaluates to negative two cos of two minus three π squared plus five.
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Okay, if we look back to the fundamental theorem of calculus, another important condition to consider is the continuity of the function lowercase π for which weβre integrating.
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Remember that the theorem states that our function must be continuous on the closed interval between π and π.
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This π and π form the limits of our integral.
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If lowercase π is not continuous over this interval, then we cannot confidently say that this relationship is true.
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To illustrate this, let us consider the function one over the square root of π₯.
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If we were to draw the graph of this function, it might look something like this.
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Now letβs consider the definite integral of our function between one and two with respect to π₯.
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This could be interpreted as the area under the curve between one and two, as shown on our graph.
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To evaluate this, we might want to reexpress one over the square root of π₯, so that the power of π₯ is in a more manageable form.
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We then use the familiar power rule of integration.
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And we would simplify.
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If we continued, weβd find no problems.
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And weβd reach numerical answer.
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Okay, what would happen instead if we were asked to evaluate the definite integral between negative one and one?
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Here, we might start to run into some problems.
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It should be clear from our graph that π of π₯ is undefined when π₯ is less than or equal to zero.
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Trying to imagine the area under our curve between these bounds would be meaningless.
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Since our function is undefined over part of the closed interval between negative one and one, it cannot be said to be continuous.
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Given this fact, thereβs no sense in continuing, since we cannot use the fundamental theorem of calculus to evaluate this integral.
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Letβs take a look at an example to illustrate this.
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Evaluate the integral between four and nine of negative two times the square root of π₯ with respect to π₯.
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For this question, weβve been asked to evaluate a definite integral.
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With questions of this type, it can sometimes be useful to move constant factors, such as the negative two, from the inside of the integrand to the outside.
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Next, we might also find it useful to reexpress our square root of π₯ as π₯ to the power of a half or π₯ to the power of 0.5.
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And weβll see why in a moment.
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To move forward with this question, weβre gonna be using the second part of the fundamental theorem of calculus.
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This gives us a way to evaluate definite integrals using the antiderivative of the function which forms the integrand.
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At this point, we note that the theorem states the function, lowercase π, must be continuous on the closed interval between π and π.
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π and π are the limits of integration, which in our case are four and nine.
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Now, the function that weβre now working with, lowercase π, is the square root of π₯, which weβve just expressed as π₯ to the power of a half.
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This function is not continuous over the entire set of real numbers, but rather is only continuous when π₯ is greater than or equal to zero.
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Luckily, both of the limits of our definite integral, four and nine, are greater than or equal to zero.
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And so we can therefore say that the square root of π₯ is continuous on the closed interval between four and nine.
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This means that itβs perfectly fine to use our theorem.
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To proceed with our evaluation, we use the power rule of integration, raising the power of π₯ by one and dividing by the new power.
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The antiderivative of π₯ to the power of half is therefore two over three times π₯ to the power of three over two.
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Again, weβre gonna shift this constant factor outside of our brackets to make our calculations easier.
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Next ,we substitute in our limits of integration.
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Now, at this stage, it might be more useful to see our power of three over two expressed as the cube of a square root.
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Conveniently, nine and four are both square numbers.
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And we can simplify our parentheses to be three cubed minus two cubed.
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We now move forward with a few more simplifications.
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And eventually, we reach an answer of negative 76 over three.
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This is the final answer to our question.
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We evaluated the given definite integral using the second part of the fundamental theorem of calculus to help us.
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Along the way, we were sure to confirm that our function lowercase π was continuous over the closed interval between the limits of integration.
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One final point that we didnβt really go into earlier.
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Weβre able to say that the square root of π₯ is not continuous when π₯ is less than zero, because, in fact, the square root of π₯ is undefined over the real numbers when π₯ is less than zero.
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And of course, a function cannot be continuous at points where it is not defined.
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Letβs now move on.
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Another thing worth considering are the cases of piecewise functions or cases involving the absolute value of a function.
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The reason we might need to think about these functions carefully is that they can be thought of as having different behaviors over different regions of their domain.
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Let us see how to deal with this in the following example.
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Evaluate the definite integral between negative four and five of the absolute value of π₯ minus two with respect to π₯.
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For this question, weβve been asked to evaluate the definite integral of a function, which weβll call lowercase π.
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This function is the absolute value or the modulus of π₯ minus two.
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Now, for any real number, we can express an absolute value function as a piecewise function.
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We can do this by recalling that if π₯ minus two evaluates to a negative number or absolute value, weβll multiply this by negative one to turn it into a positive number.
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Okay, so when π₯ minus two is greater than or equal to zero, our function is simply π₯ minus two.
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But when π₯ minus two is less than zero, our function is multiplied by negative one.
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So it is negative π₯ minus two.
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Of course, itβs probably more useful to us to isolate π₯ on one side of these inequalities.
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We do so by adding two to both sides.
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Now, it might also be useful for us to simplify this as negative π₯ plus two.
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Okay.
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Now that we have reexpressed our function piecewise, we can think about how it might look graphically.
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Here we see the graph.
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Although the scale is not exact, both the graph and the piecewise definition should show us the difference in behavior of our function either side of π₯ equals two.
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We see a sharp corner at the point two, zero on our graph.
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In fact, we would say that our function is not differentiable when π₯ equals two.
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But it is continuous when π₯ equals two.
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This is important, because in order to evaluate our definite integral, weβll be using the second part of the fundamental theorem of calculus.
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This allows us to evaluate a definite integral using the antiderivative, uppercase πΉ, of the function which forms our integrand, lowercase π.
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The condition for doing so is that lowercase π must be continuous on the closed interval between π and π, which are the limits of the integration.
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Given that our function lowercase π is continuous when π₯ is equal to two, we are able to conclude that it is continuous over the entire set of real numbers.
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And hence, the continuity condition is satisfied.
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Okay, onto evaluating the definite integral.
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Now weβve already said that our function behaves differently either side of the line π₯ equals two.
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A useful first step for us then is to split up our integral into two parts.
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The first going from the lowest bound, negative four to two, and the second going from two to five.
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Since the upper limit of our first integral is the same as the lower limit of our second integral, the sum of these two will be the same as our original integral.
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Now that weβve split our integral into two parts, weβre able to substitute in the two different subfunctions that we defined using the piecewise definition of the absolute value of π₯ minus two.
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We can understand this by considering our integrals as the area under these lines.
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From negative four to two, our function behaves like minus π₯ plus two.
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And from two to five, our function behaves like π₯ minus two.
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We can interpret the sum of these two areas as being the same as our original integral.
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From this point, we can now move forward using the familiar power rule of integration.
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We raise the power of π₯ for each of our terms and divide by the new power.
00:24:42.040 --> 00:24:45.840
Letβs tidy up to make some room for the next steps.
00:24:46.600 --> 00:24:48.800
Here, weβve input the limits of both integrals.
00:24:49.520 --> 00:24:53.600
And some color has been added to help follow the calculation.
00:24:54.440 --> 00:24:57.360
Weβll need to go through a few more simplification steps.
00:24:58.120 --> 00:25:00.480
Again, weβll clear some space.
00:25:00.800 --> 00:25:02.000
And weβll continue to simplify.
00:25:03.320 --> 00:25:08.920
Eventually, we reach a point where weβll express everything in terms of halves.
00:25:09.600 --> 00:25:16.120
And we reach a final answer of forty-five halves or 45 over two.
00:25:17.160 --> 00:25:20.880
With this, weβve completed our question.
00:25:21.520 --> 00:25:30.560
We did this by first expressing the absolute value of π₯ minus two as a piecewise function.
00:25:31.280 --> 00:25:44.120
Then, by splitting our original integral into two parts and using the second part of the fundamental theorem of calculus to help us evaluate each individual part.
00:25:45.400 --> 00:25:45.800
Okay.
00:25:46.600 --> 00:25:50.520
To finish off, letβs go through some key points.
00:25:51.600 --> 00:26:11.960
The second part of the fundamental theorem of calculus tells us if lowercase π is a continuous function on the closed interval between π and π and if capital πΉ is any antiderivative of lowercase π, which we can express as capital πΉ prime of π₯ is equal to lowercase π of π₯.
00:26:13.160 --> 00:26:22.360
Then the integral between π and π of lowercase π of π₯ with respect to π₯ is equal to capital πΉ of π minus capital πΉ of π.
00:26:23.320 --> 00:26:27.840
Remember, we can use any antiderivative of the function lowercase π.
00:26:28.840 --> 00:26:35.600
This means we can choose to take the case, which makes our calculations as simple as possible.
00:26:36.240 --> 00:26:43.120
Which is when π, our constant of integration, is equal to zero, essentially allowing us to ignore this.
00:26:44.840 --> 00:26:58.080
Often when given a definite integral, our next step will be to write the antiderivative in brackets with the limits of integration carried over to the right-hand bracket.
00:26:59.480 --> 00:27:06.080
This is a way to express the antiderivative as a function before inputting the limits of integration.
00:27:06.920 --> 00:27:12.800
But this is usually an intermediate step on your way to capital πΉ of π minus capital πΉ of π.
00:27:13.800 --> 00:27:23.600
In order to use this theorem, remember to check that lowercase π is continuous and indeed defined on the closed interval between π and π.
00:27:24.480 --> 00:27:27.240
If it is not, then the integration may fail.
00:27:28.240 --> 00:27:47.720
Finally, piecewise functions or functions involving absolute values may require you to split the integral into multiple parts, since these types of functions behave differently over different regions of their domains.