JEE MainMathematicsDefinite IntegrationMCQ+4 / −1Let f:R→Rf:R \to Rf:R→R be a differentiable function having f(2)=6f\left( 2 \right) = 6f(2)=6, f′(2)=(148)f'\left( 2 \right) = \left( {{1 \over {48}}} \right)f′(2)=(481). Then limx→2∫6f(x)4t3x−2dt\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}dt}x→2lim6∫f(x)x−24t3dt equals :A242424B363636C121212D181818Attempt in chatSharePreviousNextMore from Definite IntegrationThe value of −π∫π1+axcos2dx,a>0, is2005 · MCQIf I1=0∫12x2dx,I2=0∫12x3dx,I3=1∫22x2dx and I4=1∫22x3dx then2005 · MCQThe 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 · MCQ