Definite Integrals
8.0 Integral Inequality
8.0 Integral Inequality
(A) Let for every $x \in \left[ {a,b} \right]$
$$f\left( x \right) \le g\left( x \right) \le h\left( x \right)$$
This implies that $$\int\limits_a^b {f\left( x \right)dx} \le \int\limits_a^b {g\left( x \right)dx} \le \int\limits_a^b {h\left( x \right)dx} $$
(B) Let function ${f\left( x \right)}$ is continuous in $\left[ {a,b} \right]$ such that
Global Maximum = $M$
Global Minimum = $m$
By area under the curve, $$m\left( {b - a} \right) \le \int\limits_a^b {f\left( x \right)dx} \le M\left( {b - a} \right)$$
Question 18. Prove that $${\pi \over {128}} < \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\left( {\sin x} \right)}^{10}}dx} < {\pi \over 4}$$
Solution: For $x \in \left[ {{\pi \over 4},{\pi \over 2}} \right]$ $${1 \over {\sqrt 2 }} < \sin x < 1$$ $${\left( {{1 \over {\sqrt 2 }}} \right)^{10}} < {\left( {\sin x} \right)^{10}} < {1^{10}}$$ $$\int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{1 \over {32}}dx} < \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\left( {\sin x} \right)}^{10}}dx} < \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {1dx} $$ $${1 \over {32}}\left[ x \right]_{{\pi \over 4}}^{{\pi \over 2}} < \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\left( {\sin x} \right)}^{10}}dx} < \left[ x \right]_{{\pi \over 4}}^{{\pi \over 2}}$$ $${1 \over {32}}\left[ {{\pi \over 2} - {\pi \over 4}} \right] < \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\left( {\sin x} \right)}^{10}}dx} < \left[ {{\pi \over 2} - {\pi \over 4}} \right]$$ $${\pi \over {128}} < \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\left( {\sin x} \right)}^{10}}dx} < {\pi \over 4}$$
Question 19. Prove that $1 < \int\limits_0^{{\pi \over 2}} {{{{\mathop{\rm sinx}\nolimits} } \over x}dx} < {\pi \over 2}$
Solution: $x \in \left[ {0,{\pi \over 2}} \right]$
For $x = 0$, $${{{\mathop{\rm sinx}\nolimits} } \over x} = 1$$ For $x = {\pi \over 2}$, $${{{\mathop{\rm sinx}\nolimits} } \over x} = {2 \over \pi }$$ $${2 \over \pi } < {{{\mathop{\rm sinx}\nolimits} } \over x} < 1$$$$\int\limits_0^{{\pi \over 2}} {{2 \over \pi }dx} < \int\limits_0^{{\pi \over 2}} {{{{\mathop{\rm sinx}\nolimits} } \over x}dx} < \int\limits_0^{{\pi \over 2}} {1dx} $$ $${2 \over \pi }\left[ x \right]_0^{{\pi \over 2}} < \int\limits_0^{{\pi \over 2}} {{{{\mathop{\rm sinx}\nolimits} } \over x}dx} < \left[ x \right]_0^{{\pi \over 2}}$$ $${2 \over \pi }\left[ {{\pi \over 2} - 0} \right] < \int\limits_0^{{\pi \over 2}} {{{{\mathop{\rm sinx}\nolimits} } \over x}dx} < \left[ {{\pi \over 2} - 0} \right]$$ $$1 < \int\limits_0^{{\pi \over 2}} {{{{\mathop{\rm sinx}\nolimits} } \over x}dx} < {\pi \over 2}$$
Question 20. Prove that $2{e^{ - {1 \over 4}}} < \int\limits_0^2 {{e^{{x^2} - x}}dx} < 2{e^2}$
Solution: For $0 < x < 2$, $$ - {1 \over 4} < {x^2} - x < 2$$ $${e^{ - {1 \over 4}}} < {e^{{x^2} - x}} < {e^2}$$ $$\int\limits_0^2 {{e^{ - {1 \over 4}}}dx} < \int\limits_0^2 {{e^{{x^2} - x}}dx} < \int\limits_0^2 {{e^2}dx} $$ $${e^{ - {1 \over 4}}}\left[ x \right]_0^2 < \int\limits_0^2 {{e^{{x^2} - x}}dx} < {e^2}\left[ x \right]_0^2$$ $$2{e^{ - {1 \over 4}}} < \int\limits_0^2 {{e^{{x^2} - x}}dx} < 2{e^2}$$
(C) Let function $f\left( x \right)$ where $x \in \left[ {a,b} \right]$ then
$$\int\limits_a^b {f\left( x \right)dx} \le \left| {\int\limits_a^b {f\left( x \right)dx} } \right| \le \int\limits_a^b {\left| {f\left( x \right)} \right|dx} $$
Question 21. Let $f$ be a integrable function such that $\left| {f\left( u \right) - f\left( v \right)} \right| \le \left| {u - v} \right|$ for all real u and v in an interval $\left[ {a,b} \right]$, then prove that $$\left| {\int\limits_a^b {f\left( x \right)dx} - \left( {b - a} \right)f\left( a \right)} \right| \le {{{{\left( {b - a} \right)}^2}} \over 2}$$
Solution: $$\left| {\int\limits_a^b {f\left( x \right)dx} - \left( {b - a} \right)f\left( a \right)} \right|$$ $$ = \left| {\int\limits_a^b {f\left( x \right)dx} - \int\limits_a^b {f\left( a \right)dx} } \right|$$ $$ = \left| {\int\limits_a^b {f\left( x \right) - f\left( a \right)dx} } \right|$$ $$ \le \int\limits_a^b {\left| {f\left( x \right) - f\left( a \right)} \right|dx} $$ $$ \le \int\limits_a^b {\left| {x - a} \right|dx} $$ $$ \le \left[ {{{{{\left( {x - a} \right)}^2}} \over 2}} \right]_a^b$$ $$ \le {{{{\left( {b - a} \right)}^2}} \over 2}$$
(D) Cauchy-Schwarz Inequality
Let function ${f\left( x \right)}$ and ${g\left( x \right)}$ are continuous in $x \in \left[ {a,b} \right]$, then
$${\left[ {\int\limits_a^b {f\left( x \right)dx} } \right]^2} \le \int\limits_a^b {f{{\left( x \right)}^2}dx} \cdot \int\limits_a^b {g{{\left( x \right)}^2}dx} $$
