Find the area of the region included between the parabola y2 = x and the line x + y = 2?

Answer 1

This problem involves some tedious calculations, where a mistake can easily be made. But the following is the basic method, breaking up the total area into parts to integrate with different limits of integration.

The parabola opens to the right, going above and below the x-axis, with vertex at the origin. the line has a negative slope = -1 or 45 degrees below the horizontal, with y-intercept at (0,2) and x intercept at (2,0)

You can break up the area into parts, solve the area for each part, and add them all together.

You need the x coordinate for where the line and parabola intersect. they intersect in 2 points, one above the x-axis and one below, at (1,1) and (4,-2) It helps to sketch the graph. Draw a vertical line above the x-axis from (1,0) to (1,1) breaking up the area above the x-axis into two parts. Draw a vertical line below the x-axis from (2,0) to (2,sqr2) That line divides the area below the x-axis into two parts.

solve for y in both equations, and set them equal. y=x^1/2 and y=2-x

x^1/2 = 2-x then square both sides

x=4-4x+x² move all terms to the left side

x² - 5x +4 = 0 solve for x

factor (x-4)(x-1) set the factors = 0, solve for x x-1=0 x=1, x-4=0 x=4

x=1 is where the line and parabola intersect above the x axis. 4 is where they intersect below the x-axis. You need that to know the limits of integration. Break up the area into 4 parts, two above the x-axis and two below. with limits of integration from 0 to 1 and from 1 to 2 for above the x-axis and from 0 to 1 and 1 to 4 for below the x axis.

For above, from x= 0 to 1

The area is the integral of x^1/2 which is 2x^3/2/3 evaluated from 0 to 1, which is 2-0 = 2/3. The integral is found by increasing the power of x by 1 1/2+1=2/3 Then dividing by 2/3.. When x=1, the integral=2/3. subtract when x=0, when the integral =0 2/3 - 0 = 2/3

For above from x=1 to 2

the area is 1/2 the area of a triangle, with base=1 and height=1 or 1/2(1)(1) = 1/2

the base is the x-axis from 1 to 2, the height is from the point (1,0) to (1,1)

The total area above the x-axis is 2/3 + 1/2 = 4/6 + 3/6 = (4+3)/6 = 7/6

The area below the x-axis is in two parts, from x=0 to 2 and from x= 2 to 4

for from x=0 to 2

the area is the integral of x^1/2 or 2x^3/2/3 evaluated at 2 and 0, which is 4(2)^1/2/3 or 4/3 times square root of 2 or about 3.26

from x=2 to 4, the area is the difference of two parts

the area is the integral of x^1/2 which is 2x^3/2/3 evaluated at 4 and 2, which is 16/3 +4(2)^1/2/3

but subtract the area of a triangle formed by the line y=2-x, the x-axis and x=4. It has base 2, and height 2. It's area is half of 2 times 2 or 4/2 = 2 16/3-4(2)^1/2/3 -2 = 16/3+6/3-4(2)^1/2= 10/3+4(2)^1/2/3

Area below the x-axis is 10/3+(4/3)sqr2

Area above the x-axis is 7/6

Add them, 10/3+(4/3)sqr2 + 7/6 = 20/6+ (8/6)sqr2+7/6= 27/6 or (27+8sqr2)/6 = about 6.52 for total area between the line and parabola