Indefinite Integration 1 Question 33
34. Let $f$ be a positive function.
If $I _1=\int _{1-k}^{k} x f[x(1-x)] d x$ and
$$ I _2=\int _{1-k}^{k} f[x(1-x)] d x \text {, where } 2 k-1>0 $$
Then, $\frac{I _1}{I _2}$ is
(1997C, 2M)
(a) 2
(b) $k$
(c) $1 / 2$
(d) 1
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Solution:
- Given, $I _1=\int _{1-k}^{k} x f[x(1-x)] d x$
$$ \begin{aligned} \Rightarrow \quad I _1 & =\int _{1-k}^{k}(1-x) f[(1-x) x] d x \\ & \left.\left.=\int _{1-k}^{k} f[(1-x)] d x\right]-\int _{1-k}^{k} x f(1-x)\right] d x \\ \Rightarrow \quad I _1 & =I _2-I _1 \Rightarrow \frac{I _1}{I _2}=\frac{1}{2} \end{aligned} $$
