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Evaluate the integral from one to three of negative π₯ minus two with respect to π₯ using the limit of Riemann sums.
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Weβre asked to evaluate the definite integral of a linear function, and we know we could do this by using the power rule for integration.
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However, in this question, weβre asked to use the limit of Riemann sums.
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So letβs start by recalling how we can evaluate the limit of a definite integral by using the limit of Riemann sums.
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We recall if π is an integrable function on the closed interval from π to π, then we can evaluate the definite integral from π to π of π of π₯ with respect to π₯ by evaluating the limit as π approaches β of the sum from π equals one to π of π of π₯ π multiplied by Ξπ₯.
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And this is the limit as π approaches β of the right Riemann sum for π of π₯.
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Ξπ₯ will be our interval width.
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π minus π divided by π and π₯ π will be our sample points, π plus π times Ξπ₯.
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We want to apply this to the integral given to us in the question, the integral from one to three of negative π₯ minus two with respect to π₯.
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First, our function π of π₯ will be our integrand negative π₯ minus two.
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This means that π of π₯ is a linear function or a polynomial.
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And we know polynomials are continuous for all real values of π₯.
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And if a function is continuous on an interval, then that means it must be integrable on that interval.
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So this means our function will be integrable on any interval.
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In particular, this means that π will be integrable on any closed interval from π to π.
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Next, we can find the values of π and π.
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These will be the upper and lower limits of our integral.
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So weβll set our value of π equal to one and π equal to three.
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Weβre now ready to find an expression for Ξπ₯.
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We know Ξπ₯ is π minus π divided by π.
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We know π is three and π is one.
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So we get Ξπ₯ is three minus one divided by π, which of course simplifies to give us two divided by π.
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We can then use this to find an expression for each of our sample points π₯ π.
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We know that π is equal to one and Ξπ₯ is equal to two over π, so we get π₯ π is equal to one plus π times two over π.
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And weβll write this as one plus two π over π.
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Weβre now ready to start formulating the limit of our Riemann sum.
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We need to find an expression for π evaluated at π₯ π.
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And weβve already found an expression for π₯ π, so we need to substitute this into our definition for π of π₯.
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To evaluate π at π₯ π, we need to substitute one plus two π over π into our function π of π₯.
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Remember, π of π₯ is negative π₯ minus two.
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This gives us negative one times one plus two π over π minus two.
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And we can simplify this.
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Weβll distribute the negative one over our parentheses.
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This gives us negative one minus two π over π minus two.
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And of course, negative one minus two is equal to negative three.
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Now that weβve shown that our function π is integrable on the closed interval of our integral and we found Ξπ₯, π₯ π, and π evaluated at π₯ π, weβre ready to construct our limit.
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Substituting in π of π₯ π is negative two π over π minus three and Ξπ₯ is two over π, we get the following expression.
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The integral from one to three of negative π₯ minus two with respect to π₯ is equal to the limit as π approaches β of the sum from π equals one to π of negative two π over π minus three multiplied by two over π.
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Now, instead of evaluating the definite integral, we can instead evaluate this limit.
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Thereβs a few different ways we can approach this.
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Firstly, we could notice that both two and π are constants with respect to our sum, and this means we can just take this outside of our sum.
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Doing this gives us the limit as π approaches β of two over π times the sum from π equals one to π of negative two π over π minus three.
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At this point, we might choose to start evaluating this series.
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However, we might find it easier to represent this as two separate series.
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So, by splitting the series in two, we get the limit as π approaches β of two over π times the sum from π equals one to π of negative two π over π plus the sum from π equals one to π of negative three.
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And now we can start evaluating a little bit.
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For example, negative three is a constant.
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It doesnβt change inside of our series, so the sum from π equals one to π of negative three will just be π copies of negative three.
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Of course, this evaluates to give us negative three π.
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So evaluating our second series gave us negative three π.
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Letβs now try and evaluate our first series.
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Once again, we can see that thereβs a factor of negative two divided by π.
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This is a constant in our series, so we can just take it outside of our series.
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In other words, weβll rewrite this series as negative two over π times the sum from π equals one to π of π.
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Now, we can concentrate on this series itself, the sum from π equals one to π of π.
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These are called the triangular numbers.
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Itβs a well-known result which we can just quote.
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The sum from π equals one to π of π is equal to one-half π times π plus one.
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So we can just write this in for our series.
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This gives us the limit as π approaches β of two over π times negative two over π multiplied by one-half π times π plus one minus three π.
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And this is a very complicated-looking limit.
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However, we can simplify it to make it easier.
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First, weβll take the constant factor of two outside of our limit.
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Next, we want to distribute one over π over our outermost set of parentheses.
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In our first term, we get negative two over π multiplied by one over π is negative two over π squared.
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And in our second term, we have negative three over π multiplied by one over π.
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We can cancel the πβs to just give us negative three.
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This gives us two times the limit as π approaches β of negative two over π squared times one-half π times π plus one minus three.
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And we can then continue to simplify.
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For example, in the first term inside of our limit, we can cancel a shared factor of two in the numerator and the denominator.
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In fact, we can also cancel one of the shared factors of π in our numerator and denominator.
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Doing this, we get two times the limit as π approaches β of negative one over π times π plus one minus three.
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The last thing weβll do is distribute negative one over π over our parentheses.
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Doing this gives us two times the limit as π approaches β of negative π divided by π minus one over π minus three.
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And now weβre ready to just directly evaluate this limit.
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First, our limit is as π is approaching β.
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And since π is approaching β, π will eventually be bigger than zero.
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This means negative π divided by π is a constant and is equal to negative one.
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In our second term of negative one over π, as π is increasing, the denominator is growing without bound.
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However, our numerator remains constant.
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This means our second term limits to give us zero.
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And finally, our third term is a constant, so its limit is just itself, negative three.
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This gives us two times negative one minus three.
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And of course, two times negative one minus three is negative eight.
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And itβs also worth pointing out we can check our answer by using the power rule for integration to evaluate this definite integral.
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And if we did this, we would also get negative eight.
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Therefore, we were able to evaluate the definite integral from one to three of negative π₯ minus two with respect to π₯ by using the limit of Riemann sums.
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We got that this definite integral evaluates to give us negative eight.