Question

Draw a rough sketch of the curve y² = 4x and find the area of region enclosed by the curve and the line y = x.

Answer 1

y² = 4x is a right-handed parabola with vertex at O(0,0) and axis of parabola is x- axis.

y = x is a line passing through origin O(0,0)

Now, Finding their intersection

y² = 4x

⇒ x² = 4x

⇒ x² - 4x = 0

⇒ x(x - 4) = 0

⇒ x = 0 and x = 4

Also y = x ⇒ y = 0 and y = 4

∴ Points of intersections are (0,0) and (4,4)

Required area = `2 int_0^4 sqrt"x" "dx" - int_0^4 "x"  "dx"`

`= 2["x"^(3/2)/(3/2)]_0^4 - ["x"^2/2]_0^4`

`= 2xx2/3 |(4)^(3/2) - 0| - |4^2/2 -0 |`

`= 4/3 xx 8 - 8`

`= (32 - 24)/3 = 8/3`

`= 2 8/3 "sq.units"`