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Pattern – 31 : Definite integration

March 1, 2021
Category: Pattern – 31 : Definite integration

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Pattern – 31 :

If $f^2$ (x) and $g^2$ (x) are integrable on the interval [a, b], then prove that

$\left|\ \int_{a}^{b}f\left(x\right)\ g\left(x\right)\ dx\ \right|$ $\sqrt{\left(\int_{a}^{b}{f^2\left(x\right)\ dx}\right)\times\left(\int_{a}^{b}{g^2\left(x\right)\ dx}\right)}$

Proof :

Let, F(x) = $\left\{f\left(x\right)-xg\left(x\right)\right\}^2$ : x ∈ [0, b] ≥ 0

$\therefore\ \ \int_{a}^{b}F\left(x\right)dx\geq 0$

$\therefore\ \ \int_{a}^{b}\left\{f\left(x\right)-xg\left(x\right)\right\}^2dx\geq 0\$

$\therefore\ \ \int_{a}^{b}\left\{f^2\left(x\right)-2x\ f\left(x\right)g\left(x\right)+x^2g\left(x\right)\right\}dx\geq 0$

$\therefore\ \ x^2\int_{a}^{b}g\left(x\right)dx-2x\ \int_{a}^{b}f\left(x\right)\ g\left(x\right)dx$ + $\ \int_{a}^{b}{f^2\left(x\right)}dx\geq 0$

$\therefore\ \ x^2\int_{a}^{b}{g^2\ \left(x\right)}dx-2x\ \int_{a}^{b}f\left(x\right)g\left(x\right)\ dx$ $+\ \int_{a}^{b}{f^2\left(x\right)}dx\geq0---\left(1\right)$

∴ D ≤ 0

∴ $B^2$ – 4 AC ≤ 0 ⇒ $B^2$ ≤ 4AC

$\therefore\ \ 4\left\{\int_{a}^{b}{f\left(x\right)\ g\left(x\right)}\ dx\right\}^2\le$ $4\int_{a}^{b}{g^2\left(x\right)dx\ \int_{a}^{b}{f^2\left(x\right)}}dx$

$\therefore\ \ \left|\int_{a}^{b}f\left(x\right)g\left(x\right)dx\right|$ $\le\ \sqrt{\left(\int_{a}^{b}{f^2\left(x\right)dx}\right)\times \ \left(\int_{a}^{b}{g^2\left(x\right)dx}\right)}$

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Pattern – 31 : Definite integration

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