Indefinite Integration 3 Question 11
11. $\int _0^{x} f(t) d t=x+\int _x^{1} t f(t) d t$, then the value of $f(1)$ is
(a) $\frac{1}{2}$
(b) 0
(c) 1
(d) $-\frac{1}{2}$
(1998, 2M)
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Answer:
Correct Answer: 11. (a)
Solution:
- Given, $\int _0^{x} f(t) d t=x+\int _x^{1} t f(t) d t$
On differentiating both sides w.r.t. $x$, we get
$$ \begin{array}{rlrl} & & f(x) 1 & =1-x f(x) \cdot 1 \Rightarrow(1+x) f(x)=1 \\ \Rightarrow & & f(x)=\frac{1}{1+x} \Rightarrow f(1)=\frac{1}{1+1}=\frac{1}{2} \end{array} $$
