Let $f:[-1,2] \rightarrow[0, \infty)$ be a continuous function such that $f(x)=f(1-x), \forall x \in[-1,2]$.…
Let $f:[-1,2] \rightarrow[0, \infty)$ be a continuous function such that $f(x)=f(1-x), \forall x \in[-1,2]$. If $R_1=\int_{-1}^2 x f(x) d x$ and $R_2$ is the area of the region bounded by $y=f(x), x=-1, x=2$ and the X-asis. Then
$2 R_1=R_2$
$R_1=3 R_2$
$R_1=2 R_2$
$3 R_1=R_2$
Solution
$\left[\right.$ as $\int_a^b f(x) d x=\int_a^b f(a+b-x)$ also $\left.f(1-x)=f(x)\right]$
from (iii) and (iv) $2 R_1=R_2$
$\left[\right.$ as $\int_a^b f(x) d x=\int_a^b f(a+b-x)$ also $\left.f(1-x)=f(x)\right]$
from (iii) and (iv) $2 R_1=R_2$