The area of the region bounded by the parabola y^2 = x and the straight line 2y = x is

The area of the region bounded by the parabola $y^2 = x$y2=x and the straight line $2y = x$2y=x.

To find where the curves intersect, from the line $x = 2y$x=2y.

Substituting into the parabola equation:

$$y^2 = 2y$$y2=2y

$$y^2 - 2y = 0$$y2−2y=0

$$y(y - 2) = 0$$y(y−2)=0

So $y = 0$y=0 or $y = 2$y=2

When $y = 0$y=0: $x = 2(0) = 0$x=2(0)=0 giving point $(0, 0)$(0,0)

When $y = 2$y=2: $x = 2(2) = 4$x=2(2)=4 giving point $(4, 2)$(4,2)

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To determine which curve is on the right, check at $y = 1$y=1:

Parabola: $x = y^2 = 1$x=y2=1

Line: $x = 2y = 2$x=2y=2

Since $2 > 1$2>1, the line is to the right of the parabola between $y = 0$y=0 and $y = 2$y=2.

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Since both equations are in terms of $x$x as a function of $y$y, integrate with respect to $y$y:

$$\text{Area} = \int_0^2 (2y - y^2) \, dy$$Area=∫02​(2y−y2)dy

$$= \left[y^2 - \frac{y^3}{3}\right]_0^2$$=[y2−3y3​]02​

$$= \left[(2)^2 - \frac{(2)^3}{3}\right] - \left[0 - 0\right]$$=[(2)2−3(2)3​]−[0−0]

$$= 4 - \frac{8}{3}$$=4−38​

$$= \frac{12}{3} - \frac{8}{3}$$=312​−38​

$$= \frac{4}{3}$$=34​

The area of the region is $\frac{4}{3}$34​ square units.