Definite Integration Question 87
Question 87
- Let $f: R \rightarrow R$ 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 $I=\int_{-1}^{2} \frac{x f\left(x^{2}\right)}{2+f(x+1)} d x$, then the value of $(4 I-1)$ is
(2015 Adv.)
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Answer:
Correct Answer: 89. (0)
Solution:
- Here, $f(x)=\begin{array}{cc}{[x],} & x \leq 2 \ 0, & x>2\end{array}$
$$ \begin{aligned} & \therefore \quad I=\int_{-1}^{2} \frac{x f\left(x^{2}\right)}{2+f(x+1)} d x \ &=\int_{-1}^{0} \frac{x f\left(x^{2}\right)}{2+f(x+1)} d x+\int_{0}^{1} \frac{x f\left(x^{2}\right)}{2+f(x+1)} d x \ &+\int_{1}^{\sqrt{2}} \frac{x f\left(x^{2}\right)}{2+f(x+1)} d x+\int_{\sqrt{2}}^{\sqrt{3}} \frac{x f\left(x^{2}\right)}{2+f(x+1)} d x \ &+\int_{\sqrt{3}}^{2} \frac{x f\left(x^{2}\right)}{2+f(x+1)} d x \end{aligned} $$
$$ \begin{aligned} =\int_{-1}^{0} 0 d x+\int_{0}^{1} 0 d x+\int_{1}^{\sqrt{2}} & \frac{x \cdot 1}{2+0} d x \ & \quad+\int_{\sqrt{2}}^{\sqrt{3}} 0 d x+\int_{\sqrt{3}}^{2} 0 d x \end{aligned} $$
$\because-1<x<0 \Rightarrow 0<x^{2}<1 \Rightarrow\left[x^{2}\right]=0$,
$$ 0<x<1 \Rightarrow 0<x^{2}<1 \Rightarrow\left[x^{2}\right]=0 $$
$$ 1<x<\sqrt{2} \Rightarrow \quad 1<x^{2}<2 \quad \Rightarrow\left[x^{2}\right]=1 $$
$$ \sqrt{2}<x<\sqrt{3} \Rightarrow 2<x^{2}<3 \Rightarrow f\left(x^{2}\right)=0 $$
and $\sqrt{3}<x<2 \Rightarrow 3<x^{2}<4 \Rightarrow f\left(x^{2}\right)=0$
$$ \begin{array}{lll} \Rightarrow & I=\int_{1}^{\sqrt{2}} \frac{x}{2} d x={\frac{x^{2}}{4}}_{1}^{\sqrt{2}}=\frac{1}{4}(2-1)=\frac{1}{4} \ \therefore & 4 I=1 \Rightarrow 4 I-1=0 \end{array} $$
