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

Integrate by parts with

`u = f(\phi) \implies du = f'(\phi) \, d\phi = (2\phi - 1) \, d\phi`

`dv = d\phi \implies v = \phi`

to evaluate the integral in the definition of
`f`.

`\displaystyle \int_0^1 f(\phi) \, d\phi = \phi\,f(\phi) \bigg|_0^1 - \int_0^1 \phi(2\phi-1) \, d\phi = f(1) - \frac16`

Now if
`\phi=1`, we have

`f(1) = 1^2 - 1 - \left(f(1) - \frac16\right) \implies 2f(1) = \frac16 \implies f(1) = \frac1{12}`

and so

`\displaystyle \frac1{12} = 1^2 - 1 - \int_0^1 f(\phi) \, d\phi \implies \int_0^1 f(\phi) \, d\phi = -\frac1{12}`

It follows that

`\displaystyle \int_0^2 f(\phi) \, d\phi = \int_0^2 \left(\phi^2 - \phi + \frac1{12}\right) \, d\phi \\\\ ~~~~~~~~~~~~~~~ = \left(\frac{\phi^3}3 - \frac{\phi^2}2 + \frac\phi{12}\right) \bigg|_0^2 \\\\ ~~~~~~~~~~~~~~~ = \frac{2^3}3 - \frac{2^2}2 + \frac2{12} \\\\ ~~~~~~~~~~~~~~~ = \boxed{\frac56}`