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Question:

Let f:[-1,2]→[0,∞) be a continuous function such that f(x) = f(1-x) for all x∈[-1,2]. Let R1 = ∫_{-1}^{2} xf(x)dx, and R2 be the area of the region bounded by y=f(x), x=-1, x=2, and the x-axis. Then

R1=3R2

2R1=R2

R1=2R2

3R1=R2

Solution:

Let I = ∫{-1}^{2} xf(x)dx.
Let u = 1-x. Then du = -dx. When x = -1, u = 2. When x = 2, u = -1.
Then I = ∫{2}^{-1} (1-u)f(1-u)(-du) = ∫{-1}^{2} (1-u)f(1-u)du = ∫{-1}^{2} (1-x)f(1-x)dx.
Since f(x) = f(1-x), I = ∫{-1}^{2} (1-x)f(x)dx.
Therefore, 2I = ∫{-1}^{2} xf(x)dx + ∫{-1}^{2} (1-x)f(x)dx = ∫{-1}^{2} f(x)dx = R2.
Hence, 2R1 = R2.