Inequalities – Property IV
Property IV: If \({{f}^{2}}\left( x \right)\) and \({{g}^{2}}\left( x \right)\) are integrable on the interval [a, b]. then \(\left| \int\limits_{a}^{b}{f\left( x \right)g\left( x \right).dx} \right|\le \sqrt{\left( \int\limits_{a}^{b}{{{f}^{2}}\left( x \right).dx} \right)\left( \int\limits_{a}^{b}{{{g}^{2}}\left( x \right).dx} \right)}\).
Proof:
Let \(F\left( x \right)={{\left( f\left( x \right)-\lambda g\left( x \right) \right)}^{2}}\ge 0\),
Where \(\lambda \), is real number. Then
\(\int\limits_{a}^{b}{{{\left( f\left( x \right)-\lambda g\left( x \right) \right)}^{2}}}.dx\ge 0\),
\(\int\limits_{a}^{b}{{{f}^{2}}\left( x \right)-2\times f\left( x \right)\times \lambda g\left( x \right)+{{\lambda }^{2}}{{g}^{2}}\left( x \right)}.dx\ge 0\),
\(\int\limits_{a}^{b}{{{f}^{2}}\left( x \right).dx-2\lambda \int\limits_{a}^{b}{{}}f\left( x \right)g\left( x \right).dx+{{\lambda }^{2}}\int\limits_{a}^{b}{{}}{{g}^{2}}\left( x \right)}.dx\ge 0\),
Therefore, discriminant is non positive, i.e., \({{B}^{2}}-4AC\le 0\),
\(4{{\left( \int\limits_{a}^{b}{f\left( x \right)g\left( x \right).dx} \right)}^{2}}\le 4\int\limits_{a}^{b}{{{f}^{2}}\left( x \right).dx\int\limits_{a}^{b}{{{g}^{2}}\left( x \right)}.dx}\),
Hence
\(\left| \int\limits_{a}^{b}{f\left( x \right)g\left( x \right).dx} \right|\le \sqrt{\left( \int\limits_{a}^{b}{{{f}^{2}}\left( x \right).dx} \right)\left( \int\limits_{a}^{b}{{{g}^{2}}\left( x \right).dx} \right)}\).
Example: find the value of \(\int\limits_{0}^{1}{\sqrt{\left( 1+x \right)\left( 1+{{x}^{3}} \right)}.dx}\),
Solution:
\(\int\limits_{0}^{1}{\sqrt{\left( 1+x \right)\left( 1+{{x}^{3}} \right)}.dx}\),
\(\int\limits_{0}^{1}{\sqrt{\left( 1+x \right)\left( 1+{{x}^{3}} \right)}.dx}\le \sqrt{\left( \int\limits_{0}^{1}{\left( x+1 \right)dx} \right)\left( \int\limits_{0}^{1}{\left( \ \ \ \ \ \ \ \ 1+{{x}^{3}} \right)dx} \right)}\),
\(=\sqrt{{{\left( x+\frac{{{x}^{2}}}{2} \right)}_{0}}^{1}{{\left( x+\frac{{{x}^{4}}}{4} \right)}_{0}}^{1}}\),
\(=\sqrt{\left( 1+\frac{{{1}^{2}}}{2} \right)\left( 1+\frac{{{1}^{4}}}{4} \right)}\),
\(=\sqrt{\frac{15}{8}}\).