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6.50. Justify steps (a)–(d) to prove that if a continuous function $f$ is integrable on  an unbounded set $D$ then $\left| \int_ {D} f dA\right| \leq \int_ {D} \left|f \right|dA$    (a)$\int_ {D} f dA=\int_ {D} f_+ dA - \int_ {D} f_- dA \leq int_ \int_  {D} f_+ dA + \int_ {D} f_- dA =\int_ {D} \left|f \right|dA$   (b)$\int_ {D} (-f) dA \leq \int_ {D} \left|f \right|dA$    (c)$- \int_ {D} f dA \leq \int_ {D} \left|f \right|dA$    (d)$\left| \int_ {D} f dA\right| \leq \int_ {D} \left|f \right|dA$  (a) By Definition 6.9, if $f$ is continuous and integrable on an unbounded set $D$, then $| f |$ is integrable on $D$.   Rewrite $f(x,y) = f_+(x,y)− f_−(x,y)$ where $f_+(x,y) = f(x,y)$ if $f(x,y) \geq 0$ and 0 otherwise, and $f_−(x,y) = −f(x,y)$if $f(x,y)\leq 0$ and $0$ otherwise. So, by the definition of $\int_ {D} f dA$,   $$\int_ {D} f dA=\int_ {D} f_+ dA - \int_ {D} f_- dA $$  Since $\int_ {D} f_- dA$ is nonnegtive  $$\int_ {D} f_+ dA - \int_ {D} f_- dA \leq \int_ {D} f_+ dA + \int_ {D} f_- dA$$  Since $f_+ ≥ 0$ and $f_− ≥ 0$ are integrable over $D$