Let `F(x)=int_(x)^(x^(2)+(pi)/(6))2cos^(2)tdt` for all `x in R` and `f:[0,(1)/(2)]to[0,oo)` be continuous function. For `a in[0,(1)/(2)]`, if `F'(a)+2` is the area of the region bounded by `x=0,y=0,y=f(x)` and x = a, then f(0) is-

Let F(x)=int_x^[x^2+pi/6](2cos^2t)dt for all x in R and f:[0,1/2] -> [0,oo) be a continuous function.For a in [0,1/2] , if F'(a)+2 is the area of the region bounded by x=0,y=0,y=f(x) and x=a, then f(0) is

Let F(x)=int_(0)^(x)(cost)/((1+t^(2)))dt,0lex le2pi . Then -

Let f(x)=sin^(4)x-cos^(4)x int_(0)^((pi)/(2))f(x)dx =

Let g(x) = int_(x)^(2x) f(t) dt where x gt 0 and f be continuous function and 2* f(2x)=f(x) , then

If a continuous function f on [0, a] satisfies f(x)f(a-x)=1, a >0, then find the value of int_0^a(dx)/(1+f(x))

Let f:[-1,2]->[0,oo) be a continuous function such that f(x)=f(1-x)fora l lx in [-1,2]dot Let R_1=int_(-1)^2xf(x)dx , and R_2 be the area of the region bounded by y=f(x),x=-1,x=2, and the x-axis . Then (a) R_1=2R_2 (b) R_1=3R_2 (c) 2R_1 (d) 3R_1=R_2

If f(x)=x+int_0^1(xy^2+x^2y) dy find f(x)

Let f(x)= maximum {x^2, (1-x)^2, 2x(1 - x)} where x in [0, 1]. Determine the area of the region bounded by the curve y=f(x) and the lines y = 0,x=0, x=1.

Let f(x) =M a xi mu m {x^2,(1-x)^2,2x(1-x)}, where 0lt=xlt=1. Determine the area of the region bounded by the curves y=f(x) ,x-axis ,x=0, and x=1.

Let f(x)=int_0^x(cost)/tdt(xgt0): then at x=(2n+1)pi/2nin1, f(x) has