Let f : R → R be defined by \(\rm f(x)=\left\{\begin{matrix} \rm x+2 &\rm if\ x<0 \\ \rm |x-2| &\rm if \ x\geq0 \end{matrix}\right.\). Find \(\rm \int_{-2}^{\ \ 3} f(x)\ dx\).

Concept:

• Definite Integral: If ∫ f(x)dx = g(x) + C, then \(\rm \int_a^b f(x)\ dx = [g(x)]_a^b\) = g(b) - g(a).
• If a ≤ c ≤ b, then \(\rm \int_a^b f(x)\ dx = \int_a^c f(x)\ dx+\int_c^b f(x)\ dx\).
• \(\rm \int x^n\ dx = \frac{x^{n+1}}{n+1}+C\).

Calculation:

The given function can be summarized as follows:

Since the given function is a multi-valued function, let us separate the given definite integral into parts where the expressions of the function are different:

\(\rm \int_{-2}^{\ \ 3} f(x)\ dx=\int_{-2}^{\ \ 0} f(x)\ dx+\int_{0}^{2} f(x)\ dx+\int_{2}^{3} f(x)\ dx\)

= \(\rm \int_{-2}^{\ \ 0} (x+2)\ dx+\int_{0}^{2}(2-x)\ dx+\int_{2}^{3}(x-2)\ dx\)

= \(\rm \left[\frac{x^2}{2}+2x\right]_{-2}^{\ \ \ 0}+\left[2x-\frac{x^2}{2}\right]_{0}^{2}+\left[\frac{x^2}{2}-2x\right]_{2}^{3}\)

= \(\rm [0-(2-4)]+[4-2-0)]+\left[\frac{9}{2}-6-(2-4)\right]\)

= \(\rm 2+2+\frac{9}{2}-4\)

= 4.5