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Q.
Suppose that $ f'' (x)$ is continuous for all $x$ and $f(0) = f' (1) = 1$ If $\int \limits_0^1$ tf$^{n}$ $(t)dt=0,$ then the value of f(1) is
Integrals
Solution:
Since $\int\limits_{0}^{1} t f' \left(t\right) dt=0$
$\therefore \left|t f' \left(t\right)\right|_{0}^{1}-\int\limits_{0}^{1}1 f' \left(t\right)dt=0$
$\Rightarrow f' \left(t\right)-\int\limits_{0}^{1} f' \left(t\right)dt=0$
$\Rightarrow f' \left(1\right)-\left|f \left(t\right)\right|_{0}^{1}=0$
$\Rightarrow f' \left(1\right)-\left[f\left(1\right)-f\left(0\right)\right]=0$
$\Rightarrow f' \left(1\right)-f\left(1\right)+f\left(0\right)=0$
$\Rightarrow f\left(1\right)=f'\left(1\right)+f\left(0\right)=1+1=2$