SATHEE: Indefinite Integration 1 Question 88

Indefinite Integration 1 Question 88

89. Let $f : R \to R$ be a function defined by $f (x) = \begin{array}{cc} [x], & x \leq 2 0, & x > 2 \end{array}$ , where $[x]$ denotes the greatest integer less than or equal to $x$ . If $I = \int_{- 1}^{2} \frac{x f (x^{2})}{2 + f (x + 1)} d x$ , then the value of $(4 I - 1)$ is

(2015 Adv.)

Show Answer

Answer:

Correct Answer: 89. (0)

Solution:

Here, $f (x) = \begin{array}{cc} [x], & x \leq 2 0, & x > 2 \end{array}$

$\begin{aligned} ∴ I = \int_{- 1}^{2} \frac{x f (x^{2})}{2 + f (x + 1)} d x \\ = \int_{- 1}^{0} \frac{x f (x^{2})}{2 + f (x + 1)} d x + \int_{0}^{1} \frac{x f (x^{2})}{2 + f (x + 1)} d x \\ + \int_{1}^{\sqrt{2}} \frac{x f (x^{2})}{2 + f (x + 1)} d x + \int_{\sqrt{2}}^{\sqrt{3}} \frac{x f (x^{2})}{2 + f (x + 1)} d x \\ + \int_{\sqrt{3}}^{2} \frac{x f (x^{2})}{2 + f (x + 1)} d x \end{aligned}$

$\begin{aligned} = \int_{- 1}^{0} 0 d x + \int_{0}^{1} 0 d x + \int_{1}^{\sqrt{2}} & \frac{x \cdot 1}{2 + 0} d x \\ + \int_{\sqrt{2}}^{\sqrt{3}} 0 d x + \int_{\sqrt{3}}^{2} 0 d x \end{aligned}$

$∵ - 1 < x < 0 \Rightarrow 0 < x^{2} < 1 \Rightarrow [x^{2}] = 0$ ,

$0 < x < 1 \Rightarrow 0 < x^{2} < 1 \Rightarrow [x^{2}] = 0$

$1 < x < \sqrt{2} \Rightarrow 1 < x^{2} < 2 \Rightarrow [x^{2}] = 1$

$\sqrt{2} < x < \sqrt{3} \Rightarrow 2 < x^{2} < 3 \Rightarrow f (x^{2}) = 0$

and $\sqrt{3} < x < 2 \Rightarrow 3 < x^{2} < 4 \Rightarrow f (x^{2}) = 0$

$\begin{array}{lll} \Rightarrow & I = \int_{1}^{\sqrt{2}} \frac{x}{2} d x = {\frac{x^{2}}{4}}_{1}^{\sqrt{2}} = \frac{1}{4} (2 - 1) = \frac{1}{4} \\ ∴ & 4 I = 1 \Rightarrow 4 I - 1 = 0 \end{array}$

Indefinite Integration 1 Question 88

89. Let $f : R \to R$ be a function defined by $f (x) = \begin{array}{cc} [x], & x \leq 2 0, & x > 2 \end{array}$ , where $[x]$ denotes the greatest integer less than or equal to $x$ . If $I = \int_{- 1}^{2} \frac{x f (x^{2})}{2 + f (x + 1)} d x$ , then the value of $(4 I - 1)$ is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Indefinite Integration 1 Question 88

89. Let f:R→R be a function defined by f(x)=[x],x≤2 0,x>2, where [x] denotes the greatest integer less than or equal to x. If I=∫−12xf(x2)2+f(x+1)dx, then the value of (4I−1) is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

89. Let $f : R \to R$ be a function defined by $f (x) = \begin{array}{cc} [x], & x \leq 2 0, & x > 2 \end{array}$ , where $[x]$ denotes the greatest integer less than or equal to $x$ . If $I = \int_{- 1}^{2} \frac{x f (x^{2})}{2 + f (x + 1)} d x$ , then the value of $(4 I - 1)$ is