By the fundamental theorem of calculus,
[tex]\displaystyle\int_{-5}^5f'(x)\,\mathrm dx=f(5)-f(-5)[/tex]
The integral corresponds to the signed area under the graph of [tex]f'(x)[/tex]:
- On the interval [-5, -4], the integral is the area of a rectangle with height 5 and length 1, hence area 5*1 = 5.
- On the interval [-4, 1], the integral is the area of a right triangle with height 5 and base 5, hence area 1/2*5*5 = 25/2.
- On [1, 2], we have another triangle with height and base 1, hence area 1/2*1*1 = 1/2. But this area lies below the horizontal axis, so we take it to be negative.
- On [2, 5], we have a rectangle with height 1 and length 3, hence area 1*3 = 3, and we take it to be negative.
So, the value of the integral over [-5, 5] is
[tex]\displaystyle\int_{-5}^5f'(x)\,\mathrm dx=5+\dfrac{25}2-\dfrac12-3=14[/tex]
We're given that [tex]f(5)=-1[/tex], so
[tex]14=-1-f(-5)\implies f(-5)=\boxed{-15}[/tex]
