Class 12

Math

Calculus

Area

If the curve $y=ax_{1/2}+bx$ 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

Connecting you to a tutor in 60 seconds.

Get answers to your doubts.

The area common to regions $x_{2}+y_{2}−2x≤0andy≥2sin(πx) .$

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)(wherec∈(a,b))is minimum whenc=2a+b $. $Proof :A=a∫ c (f(c)−f(x))dx+c∫ b (f(c))dx$ $=f(c)(c−a)−a∫ c (f(x))dx+a∫ b (f(x))dx−f(c)(b−c)$ $⇒A=[2c−(a+b)]f(c)+c∫ b (f(x))dx−a∫ c (f(x))dx$ Differentiating w.r.t. c, we get $dcdA =[2c−(a+b)]f_{′}(c)+2f(c)+0−f(c)−(f(c)−0)$ For maxima and minima , $dcdA =0$ $⇒f_{′}(c)[2c−(a+b)]=0(asf_{′}(c)=0)$ Hence, $c=2a+b $ $Also forc<2a+b ,dcdA <0and forc>2a+b ,dcdA >0$ Hence, A is minimum when $c=2a+b $. If the area bounded by $f(x)=3x_{3} −x_{2}+a$ 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(1,3 1 $ 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 $A_{r}$ be the area of the region bounded between the curves $y_{2}=(e_{−kr})x(wherek>0,r∈N)and the liney=mx(wherem=0)$, k and m are some constants $A_{1},A_{2},A_{3},…$ are in G.P. with common ratio

Find the area bounded by $y_{2}≤4x,x_{2}+y_{2}≥2x,andx≤y+2$ in the first quadrant.

The value of $a(a>0)$ for which the area bounded by the curves $y=6x +x_{2}1 ,y=0,x=a,andx=2a$ has the least value is ___.