JEE Main Syllabus 2025 PDF (Out) for Physics, Chemistry, Maths

Application of Inequality in Definite Integration - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Syllabus

Quick Facts

Application of Inequality in Definite Integration is considered one of the most asked concept.
16 Questions around this concept.

Solve by difficulty

MEDIUM

Let f be a differentiable function such that $x^2 f(x)-x=4 \int_0^x t f(t) d t, f(1)=\frac{2}{3}$ Then 18 f (3) is equal to :

180

150

210

160

View Solution

Concepts Covered - 2

Application of Inequality in Definite Integration

Property 1

If f(x) ≤ g(x) ≤ h(x) for all x in [a, b], then

$\mathbf{\int_{a}^{b}f(x)\;dx\leq\int_{a}^{b}g(x)\;dx\leq\int_{a}^{b}h(x)\;dx}$

As from the figure, Area of ABDC ≤ Area of ABFE ≤ Area of ABHG

Property 2

If m is the least value (global minimum) and M is the greatest value (global maximum) of the function f(x) on the interval [a, b], then

$\mathbf{m(b-a) \leq \int_{a}^{b} f(x)\; d x \leq M(b-a)}$

Proof:

It is given that m ≤ f(x) ≤ M for all x in [a, b]

$\\\mathrm{\therefore \;\;\;\;\;\;\;\;\;\int_{a}^{b}m\;dx \leq \int_{a}^{b} f(x)\; d x \leq\int_{a}^{b} M\;dx}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;m(b-a) \leq \int_{a}^{b} f(x)\; d x \leq M(b-a)}$

Property 3

$\mathbf{\left |\int ^b_a f(x)dx \right |\leq\int^b_a \left | f(x) \right |dx}$

$\\\mathrm{We\;know\;that,\;\;\;-|f(x)|\leq f(x)\leq |f(x)|,\;\forall\;x\in[a,b]}\\\\\mathrm{\therefore \;\;\;\;\;\;\;\;\;-\int_{a}^{b}|f(x)|\;dx\leq \int_{a}^{b}f(x)\;dx\leq \int_{a}^{b}|f(x)|\;dx}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left | \int_{a}^{b} f(x)\; d x \right |\leq\int_{a}^{b}\left | f(x) \right |\;dx}$

Application of Inequality in Definite Integration (Schwarz - Bunyakovsky Inequality)

$\\\mathrm{If\;f^2(x)\;and\;g^2(x)\;are \;integrable\;function\;on\;the\;interval\;[a,b],\;then}\\\\\left|\int_{a}^{b} f(x) g(x) d x\right| \leq \sqrt{\left(\int_{a}^{b} f^{2}(x) d x\right)\left(\int_{a}^{b} g^{2}(x) d x\right).}$

Proof:

$\\\mathrm{Let\;\;F(x)=\left \{ f(x)-\lambda g(x) \right \}^2,\;\;where \;\lambda\;is\;a\;real\;number.}\\\mathrm{Since,\;\;\left \{ f(x)-\lambda g(x) \right \}^2\geq0}\\\mathrm{So,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;}\int_a^b\left \{ f(x)-\lambda g(x) \right \}^2dx\geq0\\\\\Rightarrow \mathrm{\;\;\;\;\;}\int_a^b\left \{ f^2(x)-2\lambda f(x)g(x)+\lambda^2g^2(x) \right \}dx\geq0\\\\\Rightarrow \mathrm{\;\;\;\;\;}\lambda^2\int_a^b\left \{ g^2(x) \right \}dx-2\lambda\int_a^b\left \{ f(x)g(x) \right \}dx+\int_a^b\left \{ f^2(x) \right \}dx\geq0\\\\\text{Discriminant is non positive, i.e. }\;\;B^2-4AC\leq0\\\\\Rightarrow \mathrm{\;\;\;\;}4\lambda^2\left \{ \int_a^b\left \{ f(x)g(x) \right \}dx \right \}^2\leq4\left \{ \lambda^2\int_a^b\left \{ g^2(x) \right \}dx \right \}\left \{ \int_a^b\left \{ f^2(x) \right \}dx \right \}\\\\\Rightarrow \mathrm{\;\;\;\;}\left|\int_{a}^{b} f(x) g(x) d x\right| \leq \sqrt{\left(\int_{a}^{b} f^{2}(x) d x\right)\left(\int_{a}^{b} g^{2}(x) d x\right).}$