Let f(x) be a positive function and I_1= _-(1)/(2)^12x, f(2x(1-2x)), dx and I_2= _-1^2f(x(…

Question

Let f(x) be a positive function and ${I}_{1}=\int _{-\frac{1}{2}}^{1}2x\, f(2x(1-2x))\, dx$ and ${I}_{2}=\int _{-1}^{2}f(x(1-x))\, dx$. Then the value of $\frac{{I}_{2}}{{I}_{1}}$ is equal to \_\_\_\_\_\_\_\_

Accepted Answer

$$\begin{aligned}
& I_1=\int_{-\frac{1}{2}}^1 2 x f(2 x(1-2 x)) d x \
& \Rightarrow 2 x=t \Rightarrow 2 d x=d t \quad \Rightarrow I_1=\frac{1}{2} \int_{-1}^2 t f(t(1-t)) d t \
& \Rightarrow 2 I_1=\int_{-1}^2(1-t) f(1-t)(1-(1-t)) d t
\end{aligned}$$

$$ \Rightarrow 2{I_1} = \int\limits_{ - 1}^2 {f(t(1 - t)dt - \int\limits_{ - 1}^2 {tf(t(1 - t)dt} } $$

$$\begin{aligned}
& \Rightarrow 2 \mathrm{I}_1=\mathrm{I}_2-2 \mathrm{I}_1 \
& \Rightarrow 4 \mathrm{I}_1=\mathrm{I}_2 \
& \Rightarrow \frac{\mathrm{I}_2}{\mathrm{I}_1}=4
\end{aligned}$$

Solutions · 1