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In this video, weβll look to define the definite integral of a function formally as the limit of a Riemann sum.
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In doing so, weβll establish how we can express definite integrals as limits of Riemann sums and vice versa.
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And weβll evaluate a definite integral by taking the limit of the corresponding Riemann sum written in sigma notation.
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Remember, we can estimate the area between the curve and the π₯-axis, which is bounded by the lines π₯ equals π and π₯ equals π, by splitting the region into, say, π rectangles and calculating the area of each.
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This is called finding a Riemann sum.
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And itβs defined using sigma notation as the area is approximately equal to the sum of π of π₯π star times Ξπ₯ for values of π from one to π.
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Here Ξπ₯ is equal to π minus π over π.
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Contextually, that gives us the width of each of our rectangles.
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And π₯π star is any sample point in the subinterval π₯π minus one to π₯π.
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Now, of course, it follows that as π increases, the width of each of our rectangles gets smaller.
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And this results in a more accurate estimate for the area.
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In fact, as π approaches β β the number of rectangles approaches β β the limit of this sum approaches the exact area of the region.
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We can therefore say that the area required is equal to the limit as π approaches β of the sum of π of π₯π star times Ξπ₯ for values of π from one to π.
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And in fact, this limit is hugely important as it occurs in a wide variety of situations, even when π is not a positive function.
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We therefore give it a special name, π notation.
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We come to the definition of a definite integral.
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We say that if π is a function defined for π₯ is greater than or equal to π and less than or equal to π, we can divide the closed interval π to π into π subintervals of equal width.
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We let π₯π star be the sample points in each subinterval such that π₯π star lies in the closed interval π₯π minus one to π₯π.
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Then we say that the definite integral of π from π to π is the limit as π approaches β of the sum of π of π₯π star times Ξπ₯, values of π from one to π.
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And thatβs, of course, provided that this limit exists and gives the same value for all sample points.
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If it does exist, we say that π is integrable on the closed interval π to π.
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Now this integration symbol here was introduced by Leibniz.
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Itβs an elongated S and was chosen because the integral is the limit of the sums.
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Now note that not all functions are integrable, although most commonly occurring functions are.
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In fact, if π is continuous on the closed interval π to π or if it has only a finite number of jump discontinuities, then π is integrable on the closed interval π to π.
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In other words, the definite integral from π to π of π of π₯ with respect to π₯ exists.
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Now in fact if π is integrable on the closed interval π to π, then this limit must exist.
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And it must give the same answer no matter which value, which sample point π₯π star, we choose.
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We can therefore simplify our calculation by choosing purely right π points.
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We can say that if π is integrable on the closed interval π to π, then the definite integral between π and π of π of π₯ with respect to π₯ is the limit as π approaches β of the sum of π of π₯π times Ξπ₯ for values of π from one to π.
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And here Ξπ₯ is π minus π over π, and π₯π is π plus π lots of Ξπ₯.
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This is a definition weβre going to be using throughout the remainder of this video.
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And we now have everything we need to be able to express definite integrals as limits of Riemann sums and vice versa.
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Express the definite integral between three and nine of three π₯ to the sixth power with respect to π₯ as the limit of Riemann sums.
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Remember, if π is integrable on the closed interval π to π, then the definite integral between π and π of π of π₯ with respect to π₯ can be expressed as the limit of Riemann sums as shown.
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Letβs compare everything in our theorem to our integral.
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Our function is a polynomial.
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And we know that all polynomials are continuous over their domain, which means that they are therefore integrable over their domain.
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So the function π of π₯ equals three π₯ to the sixth power is continuous and therefore integrable over the closed interval defined by the lower limit three and the upper limit nine.
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So weβre going to let π be equal to three and π be equal to nine.
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Weβll move on and define Ξπ₯.
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Itβs π minus π over π.
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Well, we said π is nine and π is three.
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And this is all over π.
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That gives us that Ξπ₯ is equal to six over π.
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And we can now define π₯π.
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Itβs π plus π lots of Ξπ₯.
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Well, π is three.
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And we need π lots of Ξπ₯, which we worked out to be six over π.
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Letβs write that as three plus six π over π.
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In our limit, weβre going to need to work out π of π₯π.
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So it follows that we can find this by substituting the expression for π₯π into our function.
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That gives us three times three plus six π over π to the sixth power.
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And we can now substitute everything we have into our definition for the integral.
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When we do, we see that the definite integral between six and nine of three π₯ to the sixth power with respect to π₯ is equal to the limit as π approaches β of the sum of three times three plus six π over π to the sixth power times six over π evaluated between π equals one and π.
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Since multiplication is commutative, we can rewrite three times six over π as 18 over π.
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And we have our definite integral expressed as the limit of Riemann sums.
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Itβs the limit as π approaches β of the sum of 18 over π times three plus six π over π to the sixth power evaluated from π equals one to π.
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Letβs now have a look at a more complicated example.
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Without evaluating the limit, express the definite integral between negative five and two of the square root of seven minus four π₯ squared with respect to π₯ as a limit of Riemann sums.
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Remember, if π is integrable on some closed interval 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 of π of π₯π times Ξπ₯ for values of π from one to π.
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We calculate Ξπ₯ by subtracting π from π and then dividing by π.
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And π₯π is π plus π lots of Ξπ₯.
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In this case, we can say that π of π₯ is equal to the square root of seven minus four π₯ squared.
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The lower limit of our integral is negative five.
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So weβll let π be equal to negative five and the upper limit is two.
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So letβs let π be equal to two.
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Itβs always sensible to next work out Ξπ₯.
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Here, thatβs π minus π.
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So thatβs two minus negative five all over π.
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That gives us Ξπ₯ to be equal to seven over π.
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Then we can work out π₯π.
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Itβs π, which is negative five, plus π lots of Ξπ₯, which we just calculated to be seven over π.
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This simplifies to negative five plus seven π over π.
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Next, we work out π of π₯π.
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And it follows that we can find this by substituting π₯π into the expression for π of π₯.
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Itβs the square root of seven minus four times negative five plus seven π over π squared.
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And we now have everything we need to be able to express our limit.
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We replace Ξπ₯ with seven over π and π of π₯π with the square root of seven minus four times negative five plus seven π over π all squared.
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And so we obtain that our integral as a limit of Riemann sums is the limit as π approaches β of the sum of seven over π times the square root of seven minus four times negative five plus seven π over π squared for values of π from one to π.
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Now this example is interesting as this equation isnβt actually integrable over the given interval.
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Remember, for π to be integrable, it must be continuous over the interval π to π.
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Well, the graph of π¦ equals the square root of seven minus four π₯ squared looks a little something like this.
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This is clearly not continuous over the closed interval negative five to two.
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So we wouldnβt be able to evaluate this limit.
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Itβs important to realise that we can carry out the procedure and get a Riemann sum.
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But we do need to check that the function is integrable over that region.
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Although we have a resulting expression in this case, it isnβt a valid answer to the question.
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If weβd instead been given limits of, say, negative square root of seven over two and positive square root of seven over two, then it wouldβve been fine.
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Letβs now see how we can reverse the process and express a limit in integral form.
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Express the limit as π approaches β of the sum of π to the power of π₯π over two minus four π₯π times Ξπ₯π for values of π from one to π as a definite integral on the closed interval negative five to negative three.
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Remember, if π is integrable on some closed interval 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 of π of π₯π times Ξπ₯ for values of π from one to π.
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Now we can quite clearly see that our interval is from negative five to negative three inclusive.
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So we begin by letting π be equal to negative five and π be equal to negative three.
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Letβs now compare our limit to the general form.
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We can see that π of π₯π is equal to π to the power of π₯π over two minus four π₯π.
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Well, thatβs great because that means π of π₯ is equal to π to the power of π₯ over two minus four π₯.
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This means the limit of our Riemann sums can be expressed as a definite integral.
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Itβs the definite integral between negative five and negative three of π to the power of π₯ over two minus four π₯ with respect to π₯.
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In our final example, weβll look at how we can actually evaluate the integral by calculating the limit of Riemann sums.
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Evaluate the definite integral between negative four and two of negative π₯ minus four with respect to π₯ using the limit of Riemann sums.
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Remember, if π is some integrable function on the closed interval π to π, then the definite integral between π and π of π of π₯ with respect to π₯ is defined as the limit as π approaches β of the sum of π of π₯π times Ξπ₯ for values of π from one to π.
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And of course, Ξπ₯ is π minus π over π and π₯π is π plus π lots of Ξπ₯.
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Letβs begin by comparing this definition with our integral.
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We see that π of π₯ is equal to negative π₯ minus four.
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This is a polynomial.
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And We know polynomials are continuous over their domain.
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So the function negative π₯ minus four is continuous and therefore integrable over the closed interval defined by the lower limit negative four and the upper limit two.
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We therefore let π be equal to negative four and π be equal to two.
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And weβll next look to define Ξπ₯.
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Itβs π minus π.
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So thatβs two minus negative four over π.
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That gives us six over π.
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Next, weβll define π₯π.
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Itβs π, which is negative four, plus π lots of Ξπ₯, which we calculated to be six over π.
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This gives us that π₯π is negative four plus six π over π.
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It follows that we can find π of π₯π by substituting this expression into our function.
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That gives us negative negative four plus six π over π minus four.
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When we distribute the parentheses, we find π of π₯π to be equal to four minus six π over π minus four.
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And of course, four minus four is zero.
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So π of π₯π is simply negative six π over π.
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And we can now substitute everything we have into our definition for the integral.
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This tells us that the definite integral between negative four and two of negative π₯ minus four with respect to π₯ is equal to the limit as π approaches β of the sum of negative six π over π times six over π for values of π from one to π.
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Now in fact, negative six π over π times six over π is negative 36π over π squared.
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So this is the sum.
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Now of course, this factor, negative 36 over π squared, is actually independent of π.
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So we can rewrite our limit as the limit as π approaches β of negative 36 over π squared times the sum of π from values of π from one to π.
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And in fact, whilst itβs outside the scope of this video to prove this, we can quote the general result.
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The sum of π from π equals one to π is equal to π times π plus one over two.
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And this means we can replace all of this sum with the expression π times π plus one over two.
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Our definite integral can therefore be evaluated by working out the limit as π approaches β of negative 36 over π squared times π times π plus one over two.
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And you might spot that we can divide both 36 and two by two.
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We can also cancel one π.
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And our limit reduces to the limit as π approaches β of negative 18 times π plus one over π.
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Letβs distribute our parentheses.
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And when we do, we find that this can be written as negative 18π over π minus 18 over π.
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Now of course, negative 18π over π is just negative 18.
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And we can now evaluate our limit by direct substitution.
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As π approaches β, negative 18 over π approaches zero.
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And so the limit as π approaches β of negative 18 minus 18 over π is simply negative 18.
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We can therefore say that the definite integral between negative four and two of negative π₯ minus four with respect to π₯ is negative 18.
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In this video, weβve seen that we can write a definite integral as the limit of Riemann sums.
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We say that, for an integrable function π over some closed interval π to π, 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 of π of π₯π times Ξπ₯ for values of π from one to π.
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Here Ξπ₯ is π minus π over π.
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And π₯π equals π plus π lots of Ξπ₯.
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We saw that we can use this definition to write an integration as a limit of a summation and vice versa.
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And that, with some clever manipulation, we can even evaluate these limits to help us calculate the integral.