Video: Evaluating the Definite Integral of a Polynomial by Taking the Limit of Riemann Sums

Evaluate ∫_(βˆ’4)^(2) (βˆ’π‘₯ βˆ’ 4) dπ‘₯ using the limit of Riemann sums.

03:46

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 𝑓 is some integrable function on the closed interval π‘Ž to 𝑏, then the definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ is defined as the limit as 𝑛 approaches ∞ of the sum of 𝑓 of π‘₯𝑖 times Ξ”π‘₯ for values of 𝑖 from one to 𝑛. And of course, Ξ”π‘₯ is 𝑏 minus π‘Ž over 𝑛 and π‘₯𝑖 is π‘Ž plus 𝑖 lots of Ξ”π‘₯.

Let’s begin by comparing this definition with our integral. We see that 𝑓 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‘Ž be equal to negative four and 𝑏 be equal to two. And we’ll next look to define Ξ”π‘₯. It’s 𝑏 minus π‘Ž. So that’s two minus negative four over 𝑛. That gives us six over 𝑛. Next, we’ll define π‘₯𝑖. It’s π‘Ž, which is negative four, plus 𝑖 lots of Ξ”π‘₯, which we calculated to be six over 𝑛. This gives us that π‘₯𝑖 is negative four plus six 𝑖 over 𝑛.

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

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 𝑛 approaches ∞ of the sum of negative six 𝑖 over 𝑛 times six over 𝑛 for values of 𝑖 from one to 𝑛. Now in fact, negative six 𝑖 over 𝑛 times six over 𝑛 is negative 36𝑖 over 𝑛 squared. So this is the sum.

Now of course, this factor, negative 36 over 𝑛 squared, is actually independent of 𝑖. So we can rewrite our limit as the limit as 𝑛 approaches ∞ of negative 36 over 𝑛 squared times the sum of 𝑖 from values of 𝑖 from one to 𝑛. And in fact, whilst it’s outside the scope of this video to prove this, we can quote the general result. The sum of 𝑖 from 𝑖 equals one to 𝑛 is equal to 𝑛 times 𝑛 plus one over two. And this means we can replace all of this sum with the expression 𝑛 times 𝑛 plus one over two.

Our definite integral can therefore be evaluated by working out the limit as 𝑛 approaches ∞ of negative 36 over 𝑛 squared times 𝑛 times 𝑛 plus one over two. And you might spot that we can divide both 36 and two by two. We can also cancel one 𝑛. And our limit reduces to the limit as 𝑛 approaches ∞ of negative 18 times 𝑛 plus one over 𝑛.

Let’s distribute our parentheses. And when we do, we find that this can be written as negative 18𝑛 over 𝑛 minus 18 over 𝑛. Now of course, negative 18𝑛 over 𝑛 is just negative 18. And we can now evaluate our limit by direct substitution. As 𝑛 approaches ∞, negative 18 over 𝑛 approaches zero. And so the limit as 𝑛 approaches ∞ of negative 18 minus 18 over 𝑛 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.

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