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 $$y=f(x)$$,$$x=$$−$$1$$ and $$x=2$$ and the $$x-axis$$. Then

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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 $$y=f(x)$$,$$x=$$−$$1$$ and $$x=2$$ and the $$x-axis$$. Then

Infinity Learn · Accepted Answer

$$R1=$$∫−$$12$$ $$xf(x)dx=$$∫−$$12$$ (2+($$−$$1)−$$x)f(2+($$−$$1)$$−$$x)dx=$$∫−$$12$$ $$(1$$−$$x)f(1$$−$$x)dx=$$∫−$$12$$ $$(1$$−$$x)f(x)dx$$⇒$$2R1=$$∫−$$12$$ $$f(x)dx=R2$$.

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 $R_{1} = \int_{- 1}^{2} x f (x) d x$ ., and $R_{2}$ be the area of the region bounded by $y = f (x), x = - 1$ and $x = 2$ and the x-axis. Then

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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 and x=2 and the x-axis. Then

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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 $R_{1} = \int_{- 1}^{2} x f (x) d x$ ., and $R_{2}$ be the area of the region bounded by $y = f (x), x = - 1$ and $x = 2$ and the x-axis. Then