Let $f:[-1,2]\to[0,\infty)$ be a continuous function such that $f(x)=f(1-x)$ for all $x\in[-1,2]$ Let $R_1=\int_{-1}^{2} xf(x)\,dx$, and $R_2$ be the area of the region bounded by $y=f(x)$, $x=-1$, $x=2$, and the $x$-axis. Then

Let f:[-1,2]→[0,∞) be a continuous function such that f(x)=f(1-x) for all x∈[-1,2] Let R₁=∫₋₁² 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 $\text{Given: } f \text{ continuous on } [-1,2],\quad f(x) = f(1-x)$ $R_1 = \displaystyle\int_{-1}^{2} xf(x)\,dx,\qquad R_2 = \int_{-1}^{2} f(x)\,dx \ (\text{area})$ $\text{To find: the relation between } R_1 \text{ and } R_2$ $1.\ \text{substitute } u = 1-x \text{ in } R_1$ $2.\ \text{use } f(1-u) = f(u)$ $3.\ \text{identify } R_2 \text{ and } R_1$ $u = 1-x,\quad x = 1-u,\quad dx = -du;\qquad x=-1 \to u=2,\ \ x=2 \to u=-1$ $R_1 = \displaystyle\int_{2}^{-1}(1-u) f(1-u)(-du) = \int_{-1}^{2}(1-u) f(u)\,du$ $R_1 = \displaystyle\int_{-1}^{2} f(u)\,du - \int_{-1}^{2} u\, f(u)\,du$ $R_1 = R_2 - R_1$ $2R_1 = R_2 \quad \Rightarrow\ \text{(3) TRUE}$ $\dfrac{R_1}{R_2} = \text{weighted average of } x = \dfrac12 \Rightarrow 2R_1 = R_2$ $\boxed{\text{Answer: (3)}}\quad 2R_1 = R_2$