Consider curves y=x21,y=4(x−1)1. Let α be the value of a(a>2) for which area bounded by curves between x=2andx=a is 1/a is e2+1andβ be the of b∈(1,2), for which the area bounded by curves between x=b and x=2 is 1−b1, then
The parabolas y2=4xandx2=4y divide the square region bounded by the lines x=4, y=4 and the coordinate axes. If S1,S2,S3 are the areas of these parts numbered from top to bottom, respectively, then
If the area bounded by the curve y=x2+1 and the tangents to it drawn from the origin is A, then the value of 3A is __.
If y=f(x) is a monotonic function in (a,b), then the area bounded by the ordinates at x=a,x=b,y=f(x)andy=f(c)(where c∈(a,b)) is minimum when c=2a+b.
Proof : A=a∫c(f(c)−f(x))dx+c∫b(f(c))dx
Differentiating w.r.t. c, we get
For maxima and minima , dcdA=0
Also for c<2a+b,dcdA<0 and for c>2a+b,dcdA>0
Hence, A is minimum when c=2a+b.
If the area enclosed by f(x)=sinx+cosx,y=a between two consecutive points of extremum is minimum, then the value of a is
If the area enclosed by the curve y=xandx=−y, the circle x2+y2=2 above the x-axis is A, then the value of π16A is__.