Answer:
(A) 6
Explanation:
since the question is about the integral and not the area of the curve, we indeed need to count the last part under the x-axis as negative.
it is an isoceles (both legs are equally long) right-angled 2×2 triangle. it's area is
2×2/2 = 2 units²
the first part is just a large rectangle 5×2, where we have to deduct two 2×1 right-angled triangles (one top left, one right).
so, we have
5×2 = 10 units²
-
2×1/2 × 2 = -2 units²
= 8 units²
and as we need to subtract the right triangle below the x-axis, the total integral is
8 - 2 = 6 units²
FYI
if the quarto was about the area under the curve, we would have to consider all parts as positron values, and we would have to calculate absolute value of the integral 0 to 5 plus the absolute value of the integral 5 to 7.