Section 5.7 : Computing Definite Integrals
17. Evaluate the following integral, if possible. If it is not possible clearly explain why it is not possible to evaluate the integral.
\[\int_{3}^{6}{{\left| {2x - 10} \right|\,dx}}\]Show All Steps Hide All Steps
We’ll need to “remove” the absolute value bars in order to do this integral. However, in order to do that we’ll need to know where \(2x - 10\) is positive and negative.
Since \(2x - 10\) is the equation of a line is should be fairly clear that we have the following positive/negative nature of the function.
\[\begin{align*}x & < 5\hspace{0.5in} \Rightarrow \hspace{0.5in}2x - 10 < 0\\ x & > 5\hspace{0.5in} \Rightarrow \hspace{0.5in}2x - 10 > 0\end{align*}\] Show Step 2So, to remove the absolute value bars all we need to do then is break the integral up at \(x = 5\).
\[\int_{3}^{6}{{\left| {2x - 10} \right|\,dx}} = \int_{3}^{5}{{\left| {2x - 10} \right|\,dx}} + \int_{5}^{6}{{\left| {2x - 10} \right|\,dx}}\]So, in the first integral we have \(3 \le x \le 5\) and so we have \(\left| {2x - 10} \right| = - \left( {2x - 10} \right)\) in the first integral. Likewise, in the second integral we have \(5 \le x \le 6\) and so we have \(\left| {2x - 10} \right| = 2x - 10\) in the second integral. Or,
\[\int_{3}^{6}{{\left| {2x - 10} \right|\,dx}} = \int_{3}^{5}{{ - \left( {2x - 10} \right)\,dx}} + \int_{5}^{6}{{2x - 10\,dx}}\] Show Step 3All we need to do at this point is evaluate each integral. Here is that work.
\[\begin{align*}\int_{3}^{6}{{\left| {2x - 10} \right|\,dx}} & = \int_{3}^{5}{{ - 2x + 10\,dx}} + \int_{5}^{6}{{2x - 10\,dx}} = \left. {\left( { - {x^2} + 10x} \right)} \right|_3^5 + \left. {\left( {{x^2} - 10x} \right)} \right|_5^6\\ & = \left[ {25 - 21} \right] + \left[ { - 24 - \left( { - 25} \right)} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{5}\end{align*}\]