Consider the function defined implicity by the equation y2−2yesin−1x+x2−1+[x]+e2sin−1x=0(where [x] denotes the greatest integer function).
The area of the region bounded by the curve and the line x=−1 is
Consider the area S0,S1,S2…. bounded by the x-axis and half-waves of the curve y=e−xsinx, where x≥0.
n=0Σ∞Sn is equal to
Area of region bounded by the curve y=416−x2 and y=sec−1[−sin2x] (where [x] denotes the greatest ingeger function) is
Two curves C1≡[f(y)]2/3+[f(x)]1/3=0andC2≡[f(y)]2/3+[f(x)]2/3=12, satisfying the relation (x−y)f(x+y)−(x+y)f(x−y)=4xy(x2−y2)
The area bounded by C1andC2 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