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Determine the integral $\int_a^e(mx+b)\,dx$ using upper and lower sums.

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Lesson Parent:

Integration III: Definite Integrals As Upper and Lower Sums

Problem:

Answer:

$\begin{equation} \int_a^e \left(mx+b\right)\,dx=\frac{m}{2}e^2+be-\left(\frac{m}{2}a^2+ba\right). \end{equation}$

Solution:

The function $mx+b$ is the formula for a straight line. In many of our integration problems we denote arbitrary limits of integration as being between $a$ and $b$ . But since the variable $b$ is in $mx+b$ , in this problem we denote arbitrary limits of integration as begin from $a$ to $e$ .

First, let's find $L(P)$ and $U(P)$ for any arbitrary partition <Sign in to see all the formulas>. (Keep in mind our notation that $a=x_0$ and $e=x_N$ .) Recall that $L(P)$ uses the minimum value of $f(x)$ in each closed interval defined by $P$ . Similarly, $U(P)$ will use the maximum value of $f(x)$ in each closed interval.

Since this is a straight line we immediately know that for each interval <Sign in to see all the formulas> that the minimum value will be <Sign in to see all the formulas> and the maximum value will be <Sign in to see all the formulas>. This gives us

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and

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Now here is the "trick" to get this to work. First, note that

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In other words, we know that the average of <Sign in to see all the formulas> and <Sign in to see all the formulas> is between the two of them. Second, note that