Question

let f:R→R be a differential equation such that f' is continuous and f(π) = -6.

if F:[0,π] → R is defined by
$\displaystyle\sf \int\limits_0^{\pi} f(t) \: dt$ , and if
$\displaystyle\sf \int\limits_0^{\pi} (f'(x) + F(x)) cosx \ dx=2$ find f(0)​

Answer 1

Answer:-

$\displaystyle\sf F(x) = \int\limits_0^x f(t) \ dt$

$\sf \implies F'(x) = f(x)$

$\displaystyle\sf I = \int\limits_0^{\pi} f'(x) \cdot cosx \ dx + \int\limits_0^{\pi} F(x) \ cos(x) \ dx = 2$

consider Value of I as (1).

let $\displaystyle\sf I_1 = \int\limits_0^{\pi} f'(x) \cdot cosx \ dx$

using by parts

$\displaystyle\sf I_1 = \bigg[ cosx \cdot f(x) \bigg]^{\pi}_0 + \int\limits_0^{\pi} sinx \cdot f(x) \ dx$

$\displaystyle\sf I_1 = 6-f(0) + \int\limits_0^{\pi} sinx \cdot F'(x) \ dx$

$\sf I_1 = 6-f(0) + I_2 \;\;\;\;\;\;\:\;\; \dots (2)$

$\displaystyle\sf I_2 = \int\limits_0^{\pi} sinx \cdot F'(x) \ dx$

again using by parts we get

$\displaystyle\sf I_2 = \bigg[ sinx \cdot F(x)\bigg]^{\pi}_0 - \int\limits_0^{\pi} cosx \cdot F(x) \ dx$

$\displaystyle\sf I_2 = - \int\limits_0^\pi cosx \cdot F(x) \ dx$

$\boxed{\displaystyle\sf (2) \implies I_1 = 6-f(0)-\int\limits_0^{\pi} cosx \cdot F(x) \ dx}$

$\boxed{\sf (1) \implies I = 6-f(0) = 2 \implies f(0) = 4}$

Answer 2

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Step-by-step explanation:

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