# Question Video: Evaluating the Definite Integral of a Polynomial by Taking the Limit of Riemann Sums Mathematics • Higher Education

Evaluate β«_(β4)^(2) (βπ₯ β 4) dπ₯ using the limit of Riemann sums.

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### Video Transcript

Evaluate the definite integral between negative four and two of negative π₯ minus four with respect to π₯ using the limit of Riemann sums.

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 Ξπ₯ for values of π from one to π. And of course, Ξπ₯ is π minus π over π and π₯π is π plus π lots of Ξπ₯.

Letβs begin by comparing this definition with our integral. We see that π of π₯ is equal to negative π₯ minus four. This is a polynomial. And We know polynomials are continuous over their domain. 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.

We therefore let π be equal to negative four and π be equal to two. And weβll next look to define Ξπ₯. Itβs π minus π. So thatβs two minus negative four over π. That gives us six over π. Next, weβll define π₯π. Itβs π, which is negative four, plus π lots of Ξπ₯, which we calculated to be six over π. This gives us that π₯π is negative four plus six π over π.

It follows that we can find π of π₯π by substituting this expression into our function. That gives us negative negative four plus six π over π minus four. When we distribute the parentheses, we find π of π₯π to be equal to four minus six π over π minus four. And of course, four minus four is zero. So π of π₯π is simply negative six π over π.

And we can now substitute everything we have into our definition for the integral. 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. Now in fact, negative six π over π times six over π is negative 36π over π squared. So this is the sum.

Now of course, this factor, negative 36 over π squared, is actually independent of π. 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. And in fact, whilst itβs outside the scope of this video to prove this, we can quote the general result. The sum of π from π equals one to π is equal to π times π plus one over two. And this means we can replace all of this sum with the expression π times π plus one over two.

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. And you might spot that we can divide both 36 and two by two. We can also cancel one π. And our limit reduces to the limit as π approaches β of negative 18 times π plus one over π.

Letβs distribute our parentheses. And when we do, we find that this can be written as negative 18π over π minus 18 over π. Now of course, negative 18π over π is just negative 18. And we can now evaluate our limit by direct substitution. As π approaches β, negative 18 over π approaches zero. And so the limit as π approaches β of negative 18 minus 18 over π is simply negative 18. 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.