Exams Joint Entrance Examination Mathematics Definite Integration Question 68 Question 68 of 246 Medium MCQ
Let the function f : [ 0 , 2 ] → R f:[0,2]\to R f : [ 0 , 2 ] → R be defined as
f ( x ) = e m i n x 2 , x − [ x ] , x ∈ [ 0 , 1 ) e [ x − log e x ] , x ∈ [ 1 , 2 ] f(x)= \begin{array}{|ccc|}{e}^{min{{x}^{2},x-[x]},} & x\in [0,1) \\ {e}^{[x-{\log }_{e}x]}, & x\in [1,2]\end{array} f ( x ) = e min x 2 , x − [ x ] , e [ x − l o g e x ] , x ∈ [ 0 , 1 ) x ∈ [ 1 , 2 ]
where [ t ] [t] [ t ] denotes the greatest integer less than or equal to t t t . Then the value of the integral ∫ 0 2 x f ( x ) d x \int _{0}^{2}xf(x)dx ∫ 0 2 x f ( x ) d x is :
Solutions · 1 Post your solution Official solution
M i n i m u m x 2 , x = x 2 ; x ∈ [ 0 , 1 ) [ x − log e x ] = 1 ; x ∈ [ 1 , 2 ) \begin{gathered} Minimum{{x}^{2},{x}} ={x}^{2};x\in [0,1) \\ \\ [x-{\log }_{e}x] =1;x\in [1,2) \end{gathered} M inim u m x 2 , x = x 2 ; x ∈ [ 0 , 1 ) [ x − log e x ] = 1 ; x ∈ [ 1 , 2 )
∴ f ( x ) = e x 2 ; x ∈ [ 0 , 1 ) e ; x ∈ [ 1 , 2 ) \therefore f(x)=\begin{gathered} {e}^{{x}^{2}};x\in [0,1) \\ e;x\in [1,2) \end{gathered} ∴ f ( x ) = e x 2 ; x ∈ [ 0 , 1 ) e ; x ∈ [ 1 , 2 )
∫ 0 2 x f ( x ) = ∫ 0 1 x ⋅ e x 2 d x + ∫ 1 2 x ⋅ e d x x 2 = t ⇒ 2 x d x = d t \begin{gathered} \int _{0}^{2}xf(x)=\int _{0}^{1}x\cdot {e}^{{x}^{2}}dx+\int _{1}^{2}x\cdot edx \\ {x}^{2}=t\Rightarrow 2xdx=dt \end{gathered} ∫ 0 2 x f ( x ) = ∫ 0 1 x ⋅ e x 2 d x + ∫ 1 2 x ⋅ e d x x 2 = t ⇒ 2 x d x = d t
= 1 2 ∫ 0 1 e t d t + e x 2 2 ∣ 2 1 =\frac{1}{2}\int _{0}^{1}{e}^{t}dt+{e\frac{{x}^{2}}{2}|}_{2}^{1} = 2 1 ∫ 0 1 e t d t + e 2 x 2 ∣ 2 1
= 1 2 ( e − 1 ) + 1 2 ( 4 − 1 ) e = 2 e − 1 2 \begin{gathered} =\frac{1}{2}(e-1)+\frac{1}{2}(4-1)e \\ =2e-\frac{1}{2} \end{gathered} = 2 1 ( e − 1 ) + 2 1 ( 4 − 1 ) e = 2 e − 2 1 Answer the question to unlock