Find the area of the region enclosed by y=−5x−x2andy=x on interval [−1,5]
Which of the following is the possible value/values of c for which the area of the figure bounded by the curves y=sin2x, the straight lines x=π/6,x=c and the abscissa axis is equal to 1/2?
The area enclosed by f(x)=12+ax±x2 coordinates axes and the ordinates at x=3(f(3)>0) is 45 sq. units. If m and n are the x-axis intercepts of the graph of y=f(x), then the value of (m+n+a) si ___.
Computing area with parametrically represented boundaries : If the boundary of a figure is represented by parametric equation, i.e., x=x(t),y=(t), then the area of the figure is evaluated by one of the three formulas :
Where αandβ are the values of the parameter t corresponding respectively to the beginning and the end of the traversal of the curve corresponding to increasing t.
If the curve given by parametric equation x=t−t3,y=1−t4 forms a loop for all values of t∈[−1,1] then the area of the loop 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 C1andx+y+2=0 is
If the area enclosed by curve y=f(x)andy=x2+2 between the abscissa x=2andx=α,α>2,is(α3−4α2+8) sq. units then find function f(x). It is known that curve y=f(x) lies below the parabola y=x2+2.