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Express the definite integral between two and five of two π₯ squared minus five over π₯ with respect to π₯ as the limit of Riemann sums.
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Remember, if π is integrable on the closed interval from π to π, then the definite integral between π and π of π of π₯ with respect to π₯ is equal to the limit as π approaches β of the sum from π equals one to π of π of π₯ π times β³π₯.
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Where β³π₯ is equal to π minus π over π and π₯ π is equal to π plus π lots of β³π₯.
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In this question, we can see that π of π₯ is equal to two π₯ squared minus five over π₯, π is the lower limit of our integral, and π is the upper limit.
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So, weβll begin by establishing whether π is actually integrable on the closed interval from π to π.
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Now, we say that if a function is continuous on a given interval, itβs also integrable on that same interval.
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So, we need to work out whether the function π of π₯ equals two π₯ squared minus five over π₯ is continuous on the closed interval from two to five.
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Well, two π₯ squared is a polynomial.
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And we know that polynomial functions are continuous over their entire domain.
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The function negative five over π₯ causes us some issues though.
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We know that when π₯ is equal to zero, five over π₯ is five divided by zero, which is undefined.
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So, the function five over π₯ is undefined at the point where π₯ equals zero.
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Luckily though, thatβs outside the interval weβre interested in.
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Weβre interested in the closed interval from two to five.
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And so, we see that five over π₯ is a rational function, which is continuous wherever it is defined.
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We also know that the sum or difference of two continuous functions is itself a continuous function.
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So, this means that π of π₯ is indeed continuous on the closed interval from two to five.
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And therefore, it must be integrable on that same interval.
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Great, so, weβve defined π to be two and π to be equal to five.
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And weβve established that π is integrable on the closed interval from two to five.
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Our next job then is to work out β³π₯.
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Itβs π minus π over π.
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So, thatβs five minus two over π, which is equal to three over π.
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Similarly, we can work out π₯ π.
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Itβs π plus π lots of β³π₯.
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So, thatβs two plus π times three over π, which is simply two plus three π over π.
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Now, of course, we need to work out π of π₯ π.
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So, thatβs π of two plus three π over π.
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We replace each instance of π₯ in our function π of π₯ with two plus three π over π.
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And when we do, we find that π of π₯ π is equal to two times two plus three π over π squared minus five over two plus three π over π.
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And weβre now ready to express our definite integral as a limit of Riemann sums.
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Itβs the limit as π approaches β of the sum from π equals one to π, so these two parts donβt change, of π of π₯ π times β³π₯, which is of course three over π.
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By convention, we bring the three over π in front of this nasty expression.
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And since weβre just being asked to write our definite integral as a limit of Riemann sums, weβre done.
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We donβt need to evaluate this.
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Itβs the limit as π approaches β of the sum from π equals one to π of three over π times two times two plus three π over π squared minus five over two plus three π over π.