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: 

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

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}

SAMPLE SOLUTION

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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

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Finally, we have

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Now we know that

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If we group the terms with $x^2$ together we have

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in the formula above. Now if we group everything with the same variable we have

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so that

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We can also group the terms with $x$ together to find that

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We can now see that

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Since we did this with an arbitrary partition, we know that

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holds for all partitions of $[a,e]$. According to our definition of a definite integral, since the number <Sign in to see all the formulas> holds for all partitions of $[a,e]$ it is <Sign in to see all the formulas>. Recall that $x_0=a$ and $x_N=e$ so that

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