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.

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