question the function f(x)=5 - |1 + x| is gra...

# Explanation: ## Step1: Split the integral interval The function \(y = 5-\vert1 + x\vert\) has a vertex at \(x=- 1\). Split the integral \(\int_{-6}^{4}(5-\vert1 + x\vert)dx\) into \(\int_{-6}^{-1}(5-( - (1 + x)))dx+\int_{-1}^{4}(5-(1 + x))dx\), which simplifies to \(\int_{-6}^{-1}(6 + x)dx+\int_{-1}^{4}(4 - x)dx\). Geometrically, the graph of \(y = 5-\vert1 + x\vert\) forms two right - angled triangles on the intervals \([-6,-1]\) and \([-1,4]\). ## Step2: Calculate the area of the first triangle For the interval \([-6,-1]\), the base of the triangle \(b_1=\vert-1-(-6)\vert = 5\), and the height \(h_1\) of the triangle at \(x=-1\) for \(y = 5-\vert1+( - 1)\vert=5\) and at \(x=-6\) for \(y = 5-\vert1+( - 6)\vert=0\), so \(h_1 = 5\). The area of a triangle \(A=\frac{1}{2}bh\), so \(A_1=\frac{1}{2}\times5\times5=\frac{25}{2}\). ## Step3: Calculate the area of the second triangle For the interval \([-1,4]\), the base of the triangle \(b_2=\vert4-( - 1)\vert=5\), and the height \(h_2\) of the triangle at \(x=-1\) for \(y = 5\) and at \(x = 4\) for \(y=5-\vert1 + 4\vert=0\), so \(h_2 = 5\). The area of the second triangle \(A_2=\frac{1}{2}\times5\times5=\frac{25}{2}\). ## Step4: Find the value of the integral The value of the definite integral \(\int_{-6}^{4}(5-\vert1 + x\vert)dx\) is the sum of the areas of the two triangles. So \(A = A_1+A_2=\frac{25}{2}+\frac{25}{2}=25\). # Answer: 25