# Calculate the area enclosed by y = X^2 and y = x + 2.

### 3 Answers | Add Yours

y= x^2

y= x+2

To find the area will need to find the integral for both equations:

First we need to determine the intersecting points:

==> x^2 = x+ 2

==> x^2 -x -2 = 0

==> (x-2)(x+1) = 0

==> x= 2 x = -1

Then we need to find the integral from x= -1 to x= 2

First let us find the area under y= x^2 from x= -1 to 2

==> A1 = intg ( x^2 ) dx

= x^3/3

= ( 2^3/3) - (-1)^3 / 3

= ( 8 + 1)/3 = 9/3 = 3

Then the area under y=x^2 from (-1 to 2 ) is 3 square units>

Now we will calculate the area under y= x+2:

A2 = intg (x+ 2)

= x^2/2 + 2x

= ( 2^2/2 + 2*2) - ( -1^2/2 + 2*-1)

= 2 + 4) - ( 1/2 - 2)

= 6 + 3/2 = 15/2 = 7.5 square units:

Then the area is:

A = A2 - A1 = 7.5 - 3 = 4.5

**Then, the area between y= x^2 and y= x+2 is 4.5 square units**

We'll calculate the area located between the given curves using Leibniz Newton formula:

S = Int (x^2 - x - 2) dx = F(b) - F(a)

Where a and b are x coordinates of the intercpeting points of the given curves.

To calculate the x coordinates of the intercepting points, we'll solve the system:

y = x^2 (1)

y = x + 2

We'll subtract 2 both sides:

x = y - 2 (2)

We'll substitute (2) in (1):

y = (y-2)^2

We'll expand the square:

y = y^2 - 4y + 4

We'll subtract y both sides and we'll use symmetric property:

y^2 - 5y + 4 = 0

y1 = 1, y2 = 4

x1 = y1 - 2

x1 = 1 - 2

x1= -1

x2 = 2

We have identified the x coordinates of the intercepting points: a = -1 and b = 2.

Now, we'll determine the expression of F(x). For this reason, we'll determine the indefinite integral of Int (x^2 - x - 2) dx using the additive property:

Int (x^2 - x - 2) dx = Int (x^2)dx - Int x dx - 2Int dx

Int (x^2 - x - 2) dx = x^3/3 - x^2/2 - 2x + C

F(-1) = -1/3 - 1/2 + 2

F(-1) = (-2-3+12)/6

F(-1) = 7/6

F(2) = 8/3 - 4/2 - 4

F(2) = (16 - 12 - 24)/6

F(2) = -10/3

S = F(2) - F(-1)

S = -10/3- 7/6

To calculate the area enclosed by y = x^2...(1) and y = x+2....(2)

We calculate the intersection points of the two curves:

Put y = x^2 in eq (2): x ^2=x+2 . Or x^2-x-2 = 0. Or (x-2)(x+1) = 0.

Therefore x -2 = 0 and x+1 = 0. Or x= 2 and y = -1. The corresponding ordinates are given by: y = x^2 . So when x= 2, y = 2^2= 4. and when x= -1, y = (-1)^2.

Therefore the curves intersect at (-1 , 1) and (2,4).

Therefore the area between (x+2) and x^2 between the ordinates ar x = -1 and x= 2 is given by:

Area = Intnt (x+2)dx - Int x^2 dx from x = -1 to 2.

Area = {x^2/2+2x-x^3/3 at x= 2} - {x^2/2 +2x -x^3/3 )atx =-1}

Area = {{4/2 +2*2 -2^3/3} - {(-1)^2/2+2(-1) - (-1)^3/3}

Area = {2+4-8/3}- {1/2 -2+1/3}

Area = 4.5

Therefore the area between f(x) = x^2 and x+2 is 4.5 sq units.