If the curve y=ax1/2+bx passes through the point (1,2) and lies above the x-axis for 0≤x≤9 and the area enclosed by the curve, the x-axis, and the line x=4 is 8 sq. units. Then
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 bounded by f(x)=3x3−x2+a and the straight lines x=0, x=2, and the x-axis is minimum, then the value of a is
The area bounded by the curve y=xe−x,y=0andx=c, where c is the x-coordinate to the curve's inflection point, is
Let O(0,0),A(2,0),andB(1,31 be the vertices of a triangle. Let R be the region consisting of all theose points P inside ΔOAB which satisfy d(P,OA)≤ min [d(P,AB)], where d denotes the distance from the point to the corresponding line. Sketch the region R and find its area.
The positive valu of the parameter 'k' for which the area of the figure bounded by the curve y=sin(kx),x=3k2π,x=3k5π and x-axis is less than 2 can be
Let Ar be the area of the region bounded between the curves y2=(e−kr)x(where k>0,r∈N) and the line y=mx(where m=0), k and m are some constants
A1,A2,A3,… are in G.P. with common ratio