Let f(x) be a positive function and $I_{1} = \int\limits_{-\frac{1}{2}}^{1} 2x \, f(2x(1-2x)) \, dx$ and $I_{2} = \int\limits_{-1}^{2} f(x(1-x)) \, dx$. Then the value of $\frac{I_{2}}{I_{1}}$ is equal to ________
Solution
<p>$$\begin{aligned}
& I_1=\int_{-\frac{1}{2}}^1 2 x f(2 x(1-2 x)) d x \\
& \Rightarrow 2 x=t \Rightarrow 2 d x=d t \quad \Rightarrow I_1=\frac{1}{2} \int_{-1}^2 t f(t(1-t)) d t \\
& \Rightarrow 2 I_1=\int_{-1}^2(1-t) f(1-t)(1-(1-t)) d t
\end{aligned}$$</p>
<p>$$ \Rightarrow 2{I_1} = \int\limits_{ - 1}^2 {f(t(1 - t)dt - \int\limits_{ - 1}^2 {tf(t(1 - t)dt} } $$</p>
<p>$$\begin{aligned}
& \Rightarrow 2 \mathrm{I}_1=\mathrm{I}_2-2 \mathrm{I}_1 \\
& \Rightarrow 4 \mathrm{I}_1=\mathrm{I}_2 \\
& \Rightarrow \frac{\mathrm{I}_2}{\mathrm{I}_1}=4
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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