Find the area enclosed by the parabola 4y=3x2 and the line 2y=3x+12
The area enclosed between the parabola,4y=3x2,and the line,2y=3x+12,is
represented by the shaded area OBAO as
The points of intersection of the given curves are A(–2,3)and(4,12).
We draw AC and BD perpendicular to x-axis.
∴Area OBAO=Area CDBA–(Area ODBO+Area OACO)
=
\[\int_{-2}^{1} \frac12(3x+12) \,dx\]\[-\int_{-2}^{1} \frac{3x^2}{4} \,dx\]=\(\frac12\)[\(\frac{3x^2}{2}\)+12x]4-2-\(\frac34\)[x3/3]4-2
=\(\frac12\)[24+48-6+24]-\(\frac14\)[64+8]
=\(\frac12\)[90]-\(\frac14\)72]
=45-18
=27units.
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Using integration finds the area of the region bounded by the triangle whose vertices are (–1, 0),(1, 3)and(3, 2).
Using integration find the area of the triangular region whose sides have the equations y =2x+1,y=3x+1 and x=4.