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Prove that if $f(x)$ is continuous on $[a,b]$ and we have a point $e$ such that $a\lt e\lt b$ , then $\begin{eqnarray}&&\int_a^b\,f(x)dx\\&=&\int_a^e\,f(x)dx+\int_e^b\,f(x)dx.\end{eqnarray}$

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Lesson Parent:

Integration III: Definite Integrals As Upper and Lower Sums

Problem:

Prove that if $f(x)$ is continuous on $[a,b]$ and we have a point $e$ such that $a\lt e\lt b$ , then
$\begin{equation} \int_a^b\,f(x)dx=\int_a^e\,f(x)dx+\int_e^b\,f(x)dx. \end{equation}$

Answer:

It is true that if $f(x)$ is continuous on $[a,b]$ and we have a point $e$ such that $a\lt e\lt b$ , then $\begin{equation} \int_a^b\,f(x)dx=\int_a^e\,f(x)dx+\int_e^b\,f(x)dx. \end{equation}$

Solution:

What we are wanting to prove is that the area under a function $f(x)$ can be broken into pieces. The area of the whole is the sum of the areas broken into two pieces:

Recall our definition of a definite integral using upper and lower sums. In that definition

$P$ to create partition

$Q$ . Furthermore, we know that

<Sign in to see all the formulas>

Now let's recall that

<Sign in to see all the formulas>

Since the partition

$Q$ contains the point

$e$ , we can break

$L(Q)$ into two parts. One part where the intervals in the partition are in

$[a,e]$ and the part where the intervals in the partition are in

$[e,b]$ . Let's say that the point

$x_j=e$ so we can write

<Sign in to see all the formulas>

where

$Q_a$ is a partition with all the points of

$Q$ in

$[a,e]$ and

$Q_b$ is a partition with all the points of

$Q$ in

$[e,b]$ . (Math afficionados would say <Sign in to see all the formulas> and <Sign in to see all the formulas>.) We also find the same thing for the upper sum

$U(Q)$ so that

Prove that if f(x)f(x) is continuous on [a,b][a,b] and we have a point ee such that a<e<ba\lt e\lt b, then ∫baf(x)dx=∫eaf(x)dx+∫bef(x)dx.\begin{eqnarray}&&\int_a^b\,f(x)dx\\&=&\int_a^e\,f(x)dx+\int_e^b\,f(x)dx.\end{eqnarray}

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SAMPLE SOLUTION

GAIN AN ADVANTAGE

Prove that if $f(x)$ is continuous on $[a,b]$ and we have a point $e$ such that $a\lt e\lt b$ , then $\begin{eqnarray}&&\int_a^b\,f(x)dx\\&=&\int_a^e\,f(x)dx+\int_e^b\,f(x)dx.\end{eqnarray}$