a) To find the coordinates of the points P and Q, we need to solve the equations y=3x and y=4-x^2 simultaneously.
Substitute y=3x into y=4-x^2:
3x = 4- x^2
Rearranging the equation, we get x^2 + 3x - 4 = 0
(x+4)(x-1) = 0
x = -4 or x = 1
When x = -4, y = 3(-4) = -12
When x = 1, y = 3(1) = 3
Therefore, point P is (-4, -12) and point Q is (1, 3).
b)
i) The area bounded by the curve y=4-x^2, the x-axis, and the line QN can be found by integrating from x=1 to x=-1 (since N is on the x-axis).
Area = ∫[1,-1] (4-x^2) dx
= [4x - (x^3)/3] from 1 to -1
= (-4 + (1/3)) - (4 - (1/3))
= -8/3 + 12/3
= 4/3 square units
ii) Since the shaded region lies below the x-axis, the area can be found by integrating from x=-1 to x=-4.
Area = ∫[-1,-4] (4-x^2) dx
= [4x - (x^3)/3] from -1 to -4
= (-16 + 4/3) - (-4 + 1/3)
= -48/3 + 4/3
= -44/3 square units
iii) The area of the region enclosed by the curve y=4-x^2, the line y=3x, and the y-axis can be found by integrating from x=-4 to x=1.
Area = ∫[-4,1] (4-x^2 - 3x) dx
= [4x - (x^3)/3 - (3x^2)/2] from -4 to 1
= (4 - (1/3) - (3/2)) - (-16 + (64/3) - (48/2))
= 4 - 1/3 - 3/2 + 16 - 64/3 + 24
= 37 - 1/3 - 3/2 -64/3
= 99/6
= 16.5 square units
A diagram shows a sketch of the line y=3x and the curve y=4-x^2 intersecting at point P and Q
a)Find the coordinates of P and Q
b)Given that QN is perpendicular to the x-axis at N,calculate;
i)the area bounded by the curve y=4-x^2,the x-axis and the line QN
ii)the area of the shaded region that lies below the x-axis
iii)the area of the region enclosed by the curve y=4-x^2,the line y=3x and the y-axis
1 answer