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Show that if we are given two partitions ($P$ and $Q$) that differ by a single point so that $P\subset Q$ then $L(P)\leq L(Q)\leq\int_a^b\,f(x)dx\leq U(Q)\leq U(P)$.

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Lesson Parent:

Integration III: Definite Integrals As Upper and Lower Sums

Problem:

Answer:

It is true that if we are given two partitions ($P$ and $Q$) that differ by a single point so that $P\subset Q$ then $L(P)\leq L(Q)\leq\int_a^b\,f(x)dx\leq U(Q)\leq U(P)$.

Solution:

Recall that we chose a partition to approximate a definite integral using upper and lower sums. In this problem we are wanting to see what the affect of adding just one single point to an existing partition will have on the evaluation of upper and lower sums of a definite integral.

Let's call our existing partition <Sign in to see all the formulas> and let's add a single point between <Sign in to see all the formulas> and $x_i$. Let's call that point $x_j$ and our new partition $Q$ so that

<Sign in to see all the formulas>

With this new partition we will find $L(Q)$ and $U(Q)$. So our question is what is $L(Q)$ and $U(Q)$ in relation to $L(P)$ and $U(P)$? What we will show is that

<Sign in to see all the formulas>

In other words, by adding only a single point to the partition $P$ we can get closer to <Sign in to see all the formulas>.

Recall that

<Sign in to see all the formulas>

Let's call all the parts of this sum except <Sign in to see all the formulas> $C$ so that <Sign in to see all the formulas>.

Now let's turn our attention to <Sign in to see all the formulas> and establish some notation. We know that <Sign in to see all the formulas>. In our new partition $Q$ we will have this broken up into two pieces. In this case

<Sign in to see all the formulas>

The variables $m'_j$ and $m'_i$ are the minimum values of $f(x)$ in the closed intervals <Sign in to see all the formulas> and <Sign in to see all the formulas>, respectively.

Now because $m_i$ is the minimum value in the interval <Sign in to see all the formulas> and because both <Sign in to see all the formulas> and <Sign in to see all the formulas> are in <Sign in to see all the formulas> we know that

<Sign in to see all the formulas>

Now using our rules of inequalities we know that:

<Sign in to see all the formulas>

and that

<Sign in to see all the formulas>

Now let's do some algebra on the first equation above:

<Sign in to see all the formulas>

Now let's do

<Sign in to see all the formulas>

But from above we have

<Sign in to see all the formulas>

So we have shown that <Sign in to see all the formulas>. Now let's add $C$ to both sides to get:

<Sign in to see all the formulas>

So we have now shown that by adding a single point to the partition $P$ we obtain a new partition $Q$ where <Sign in to see all the formulas>.

If you understand everything that we have done up to this point then the rest of the proof is basically the same thing except for the upper sums. In other words, if you understand the above the the rest will be a bit monotonous. If you don't understand the above then don't go any further until you understand the stuff above. So now let's keep going to show the details for the upper bounds.

So now let's look at $U(P)$ and $U(Q)$. We will take the same approach and notation where we put $x_j$ in between <Sign in to see all the formulas> and $x_i$ and denote $C$ by those parts of the sum that are not affected by the new point. In this case we have

Show that if we are given two partitions ($P$ and $Q$) that differ by a single point so that $P\subset Q$ then $L(P)\leq L(Q)\leq\int_a^b\,f(x)dx\leq U(Q)\leq U(P)$.

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SAMPLE SOLUTION

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