JEE MainMathematicsDefinite IntegrationMCQ+4 / −1Let the function f:[0,2]→Rf:[0,2] \rightarrow \mathbb{R}f:[0,2]→R be defined as f(x)={emin{x2,x−[x]},x∈[0,1)e[x−logex],x∈[1,2]f(x)= \begin{cases}e^{\min \left\{x^{2}, x-[x]\right\},} & x \in[0,1) \\ e^{\left[x-\log _{e} x\right]}, & x \in[1,2]\end{cases}f(x)={emin{x2,x−[x]},e[x−logex],x∈[0,1)x∈[1,2] where [t][t][t] denotes the greatest integer less than or equal to ttt. Then the value of the integral ∫02xf(x)dx\int_{0}^{2} x f(x) d x∫02xf(x)dx is :A2e−12 e-12e−1B2e−122 e-\frac{1}{2}2e−21C1+3e21+\frac{3 e}{2}1+23eD(e−1)(e2+12)(e-1)\left(e^{2}+\frac{1}{2}\right)(e−1)(e2+21)Attempt in chatSharePreviousNextMore from Definite IntegrationIf ∫−0.150.15100x2−1dx=3000k, then k is equal to .2023 · Numerical∫0∞e3x+6e2x+11ex+66dx=2023 · MCQLet for x∈R,S0(x)=x,Sk(x)=Ckx+k∫0xSk−1(t)dt, where C0=1,Ck=1−∫01Sk−1(x)dx,k=1,2,3,… Then S2(3)+6C3 is equal to .2023 · NumericalThe value of 0∫4πe−x(tan49x+tan51x)dxe−4π+0∫4πe−xtan50xdx is2023 · MCQLet fn=∫02π(∑k=1nsink−1x)(∑k=1n(2k−1)sink−1x)cosxdx,n∈N. Then f21−f20 is equal to 2023 · NumericalIf 0∫1(5+2x−2x2)(1+e(2−4x))1dx=α1loge(βα+1),α,β>0, then α4−β4 is equal to :2023 · MCQThe value of 120∫3x2−3x+2dx is 2023 · NumericalThe value of π80∫2π(sinx)2023+(cosx)2023(cosx)2023dx is 2023 · Numerical