Question

If given that :-

${\quad \leadsto \quad \bf f(\phi) = {\phi}^{2} - \phi - \displaystyle \bf \int_{\bf 0}^{1} f(\phi) d \phi }$

Then find :-

${ \quad \leadsto \quad \displaystyle \bf \int_{0}^{2} f(\phi) d \phi }$ ​

Answer 1

Answer:

int (0)^(pi) [f(x)+f''(x)]sinx dx

= int (0)^(pi)f(x)sinx dx+int (0)^(pi) f''(x)sin x dx

= (f(x)(-cosx)) (0)^(pi)+int (0)^(pi)f'(x)cosxdx

+sinxf'(x)|_(0)^(pi)-int_(0)^(pi) cos xf'(x)dx

= f(pi) + f(0) = 5. (given).

:. f(0)=5 - f(pi) = 5 - 2 = 3.

ease mark me as Brainlist

Answer 2

Answer:

int (0)^(pi) [f(x)+f''(x)]sinx dx

= int (0)^(pi)f(x)sinx dx+int (0)^(pi) f''(x)sin x dx

= (f(x)(-cosx)) (0)^(pi)+int (0)^(pi)f'(x)cosxdx

+sinxf'(x)|_(0)^(pi)-int_(0)^(pi) cos xf'(x)dx

= f(pi) + f(0) = 5. (given).

:. f(0)=5 - f(pi) = 5 - 2 = 3.

[tex]\large\bf{Hope~ Helps:)}[/tex'