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### Section 5-7 : Computing Definite Integrals

18. Evaluate the following integral, if possible. If it is not possible clearly explain why it is not possible to evaluate the integral.

$\int_{{ - 1}}^{0}{{\left| {4w + 3} \right|\,dw}}$

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Hint : In order to do this integral we need to “remove” the absolute value bars from the integrand and we should know how to do that by this point.
Start Solution

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 $$4w + 3$$ is positive and negative.

Since $$4w + 3$$ is the equation a line is should be fairly clear that we have the following positive/negative nature of the function.

\begin{align*}w & < - \frac{3}{4}\hspace{0.5in} \Rightarrow \hspace{0.5in} 4w + 3 < 0\\ w & > - \frac{3}{4}\hspace{0.5in} \Rightarrow \hspace{0.5in} 4w + 3 > 0\end{align*} Show Step 2

So, to remove the absolute value bars all we need to do then is break the integral up at $$w = - \frac{3}{4}$$.

$\int_{{ - 1}}^{0}{{\left| {4w + 3} \right|\,dw}} = \int_{{ - 1}}^{{ - \frac{3}{4}}}{{\left| {4w + 3} \right|\,dw}} + \int_{{ - \frac{3}{4}}}^{0}{{\left| {4w + 3} \right|\,dw}}$

So, in the first integral we have $$- 1 \le w \le - \frac{3}{4}$$ and so we have $$\left| {4w + 3} \right| = - \left( {4w + 3} \right)$$ in the first integral. Likewise, in the second integral we have $$- \frac{3}{4} \le w \le 0$$ and so we have $$\left| {4w + 3} \right| = 4w + 3$$ in the second integral. Or,

$\int_{{ - 1}}^{0}{{\left| {4w + 3} \right|\,dw}} = \int_{{ - 1}}^{{ - \frac{3}{4}}}{{ - \left( {4w + 3} \right)\,dw}} + \int_{{ - \frac{3}{4}}}^{0}{{4w + 3\,dw}}$ Show Step 3

All we need to do at this point is evaluate each integral. Here is that work.

\begin{align*}\int_{{ - 1}}^{0}{{\left| {4w + 3} \right|\,dw}} & = \int_{{ - 1}}^{{ - \frac{3}{4}}}{{ - 4w - 3\,dw}} + \int_{{ - \frac{3}{4}}}^{0}{{4w + 3\,dw}} = \left. {\left( { - 2{w^2} - 3w} \right)} \right|_{ - 1}^{ - \frac{3}{4}} + \left. {\left( {2{w^2} + 3w} \right)} \right|_{ - \frac{3}{4}}^0\\ & = \left[ {\frac{9}{8} - 1} \right] + \left[ {0 - \left( { - \frac{9}{8}} \right)} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{5}{4}}}\end{align*}