13. y = 4 - x, 0 ≤ x ≤ 6\n14. y = cos x, 0 ≤ x ≤ π

Explanation:

Step1: Determine where the function crosses the x - axis

Set (y = 4 - x=0), then (x = 4).

Step2: Split the integral based on the sign of the function

For (y = 4 - x) on the interval ([0,6]), we split the integral into two parts: (\int_{0}^{4}(4 - x)dx+\int_{4}^{6}-(4 - x)dx). For (\int_{0}^{4}(4 - x)dx), use the power - rule (\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)): (\int_{0}^{4}(4 - x)dx=\int_{0}^{4}4dx-\int_{0}^{4}xdx=4x|{0}^{4}-\frac{x^{2}}{2}|{0}^{4}=(4\times4 - 0)-(\frac{4^{2}}{2}-0)=16 - 8 = 8). For (\int_{4}^{6}-(4 - x)dx=\int_{4}^{6}(x - 4)dx=\int_{4}^{6}xdx-\int_{4}^{6}4dx=\frac{x^{2}}{2}|{4}^{6}-4x|{4}^{6}=(\frac{6^{2}}{2}-\frac{4^{2}}{2})-(4\times6 - 4\times4)=(\frac{36}{2}-\frac{16}{2})-(24 - 16)=(18 - 8)-8 = 2).

Step3: Calculate the total area

The total area (A=8 + 2=10).

Answer:

10