Find the area bounded by the curve , x = 2y + 3 in the first quadrant and x-axis

y = √x

Squaring both sides

⇒ y² = x

In y² = x parabola it is not defined for negative values of x hence the parabola will be to the right of Y-axis passing through (0, 0)

But we have to plot y = √x which means x,y both can only be positive hence the graph has to be only in 1^st quadrant

Hence y = √x will be part of parabola y² = x only above X-axis in 1^st quadrant

(why can’t y be negative? Because the symbol square root itself denotes primary root which means positive root)

x = 2y + 3 is a straight line

To get the intersection points of the straight line and the parabola solve the parabola equation and straight line equation simultaneously

Put x = 2y + 3 in y² = x

⇒ y² = 2y + 3

⇒ y² – 2y – 3 = 0

⇒ y² – 3y + y – 3 = 0

⇒ y(y – 3) + 1(y – 3) = 0

⇒ (y + 1)(y – 3) = 0

⇒ y = -1 and y = 3

Put y = 3 in y² = x

⇒ x = 3²

⇒ x = 9

y = -1 doesn’t matter because we are in 1^st quadrant

Hence the parabola and straight line intersects at (9, 3)

Plot roughly the parabola and the line and the required area is the shaded region as shown

The point of intersection of straight line with X-axis can be calculated by putting y = 0 in line equation

⇒ x = 2(0) + 3 ⇒ x = 3

Observe that

⇒ area bounded = area under y = √x – area under straight line …(i)

Let us fling area under y = √x

Integrate from 0 to 9

Now area under the straight line x = 2y + 3

Integrate from 3 to 9

(why 3 to 9? Because the line cuts X-axis at 3 and the curve y = √x at 9)

Using (i)

⇒ area bounded = 18 – 9 = 9 unit²

Hence area bounded by curve and straight line is 9 unit²