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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.