JEE MainMathematicsDefinite IntegrationMCQ+4 / −1If I1=∫012x2dx,I2=∫012x3dx, I3=∫122x2dx{I_1} = \int\limits_0^1 {{2^{{x^2}}}dx,{I_2} = \int\limits_0^1 {{2^{{x^3}}}dx,\,{I_3} = \int\limits_1^2 {{2^{{x^2}}}dx} } }I1=0∫12x2dx,I2=0∫12x3dx,I3=1∫22x2dx and I4=∫122x3dx{I_4} = \int\limits_1^2 {{2^{{x^3}}}dx}I4=1∫22x3dx thenAI2>I1{I_2} \gt {I_1}I2>I1BI1>I2{I_1} \gt {I_2}I1>I2CI3=I4{I_3} = {I_4}I3=I4DI3>I4{I_3} \gt {I_4}I3>I4Attempt in chatSharePreviousNextMore from Definite IntegrationThe value of integral, 3∫69−x+xxdx is2005 · MCQThe value of I=0∫π/21+sin2x(sinx+cosx)2dx is2004 · MCQThe value of −2∫31−x2dx is2004 · MCQIf 0∫πxf(sinx)dx=A0∫π/2f(sinx)dx, then A is2004 · MCQIf f(x)=1+exex,I1=f(−a)∫f(a)xg{x(1−x)}dx and I2=f(−a)∫f(a)g{x(1−x)}dx,…2004 · MCQIf f(y)=ey,g(y)=y;y>0 and F(t)=0∫tf(t−y)g(y)dy, then :2003 · MCQThe value of x→0limxsinx0∫x2sec2tdt is2003 · MCQLet f(x) be a function satisfying f′(x)=f(x) with f(0)=1 and g(x) be a function that satisfies f(x)+g(x)=x2. Then the value of the integral 0∫1f(x)g(x)dx,…2003 · MCQ