let-f-C-0-R-R-x-R-f-a-b-x-f-x-1-find-a-b-xf-x-dx-interms-of-a-b-f-x-dx-2-calculate-0-pi-xdx-1-sinx-

Question Number 50413 by Abdo msup. last updated on 16/Dec/18

l e t f \in C^{0} (R, R) / \forall x \in R f (a + b - x) = f (x)

1) f i n d \int_{a}^{b} x f (x) d x i n t e r m s o f \int_{a}^{b} f (x) d x

2) c a l c u l a t e \int_{0}^{π} \frac{x d x}{1 + s i n x}

Commented by Abdo msup. last updated on 24/Dec/18

1) \int_{a}^{b} x f (x) d x = \int_{a}^{b} x f (a + b - x) d x

=_{a + b - x = t} - \int_{b}^{a} (a + b - t) f (t) d t

= \int_{a}^{b} (a + b) f (t) - \int_{a}^{b} t f (t) d t \Rightarrow

2 \int_{a}^{b} x f (x) d x = (a + b) \int_{a}^{b} f (t) d t \Rightarrow

\int_{a}^{b} x f (x) d x = \frac{a + b}{2} \int_{a}^{b} f (x) d x

2) w e h a v e \int_{0}^{π} \frac{x d x}{1 + s i n x} = \int_{0}^{π} x f (x) d x w i t h

f (x) = \frac{1}{1 + s i n x} w e h a v e f (0 + π - x) = \frac{1}{1 + s i n (π - x)}

= f (x) \Rightarrow \int_{0}^{π} x f (x) d x = \frac{π}{2} \int_{0}^{π} \frac{d x}{1 + s i n x}

=_{t a n (\frac{x}{2}) = t} \frac{π}{2} \int_{0}^{\infty} \frac{1}{1 + \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

= π \int_{0}^{\infty} \frac{d t}{1 + t^{2} + 2 t} = π \int_{0}^{\infty} \frac{d t}{{(t + 1)}^{2}}

= π {[- \frac{1}{t + 1}]}_{0}^{+ \infty} = π .