area y=2x^2,y=2x+6 — step-by-step solution

1) Consider the two curves with equations

y = 2 x^{2}

and

y = 2 x + 6

At the intersections,

2 x^{2}

2 x + 6

. Subtract the second expression from the first:

2 x^{2} - (2 x + 6)

= 0, giving

2 x^{2} - 2 x - 6 = 0

Let's solve in

ℝ

2 x^{2} - 2 x - 6 = 0

Since

2 \neq 0

, divide both sides by

2

x^{2} - x - 3 = 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -1

and

c = -3

Δ = {(-1)}^{2} - 4 \cdot 1 \cdot (-3)

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

{(-1)}^{2}

(-1) \cdot (-1)

1

Δ = 1 - 4 \cdot (-3)

ⓘ Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

- 4 \cdot (-3)

12

Δ = 1 + 12

ⓘ Start from

1

and move forward

12

on the number line:

1 + 12

13

Δ = 13

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-1) - \sqrt{13}}{2 \cdot 1}

x_{1} = \frac{1 - \sqrt{13}}{2}

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-1) + \sqrt{13}}{2 \cdot 1}

x_{2} = \frac{1 + \sqrt{13}}{2}

Solutions found:

x = {\frac{1 - \sqrt{13}}{2}, \frac{1 + \sqrt{13}}{2}}

The area between

2 x^{2}

and

2 x + 6

[\frac{1 - \sqrt{13}}{2}, \frac{1 + \sqrt{13}}{2}]

is the integral of the absolute value of their difference

No interior intersection: the difference keeps its sign on the interval

[\frac{1 - \sqrt{13}}{2}, \frac{1 + \sqrt{13}}{2}]

2 x + 6

is above

2 x^{2}

: integrate

- 2 x^{2} + 2 x + 6

Between its roots

x_{1} = \frac{1 - \sqrt{13}}{2}

and

x_{2} = \frac{1 + \sqrt{13}}{2}

- 2 x^{2} + 2 x + 6

-2 \cdot (x - x_{1}) \cdot (x - x_{2})

With

u = x - x_{1}

and

d = x_{2} - x_{1}

, the integral of

(x - x_{1}) \cdot (x - x_{2})

from

x_{1}

x_{2}

is that of

u \cdot (u - d)

from 0 to

d

\frac{d^{3}}{3} - \frac{d^{3}}{2} = - \frac{d^{3}}{6}

, so the area is

\frac{2 d^{3}}{6}

Compute

d = x_{2} - x_{1}

⇥

\frac{1 + \sqrt{13}}{2} - \frac{1 - \sqrt{13}}{2}

⇥ⓘ The minus sign of the fraction goes into the numerator or the denominator, changing the sign of each of its terms:

- \frac{1 - \sqrt{13}}{2}

\frac{- 1 + \sqrt{13}}{2}

⇥

\frac{1 + \sqrt{13}}{2} + \frac{- 1 + \sqrt{13}}{2}

⇥ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{1 + \sqrt{13}}{2} + \frac{- 1 + \sqrt{13}}{2}

\frac{(1 + \sqrt{13}) + (- 1 + \sqrt{13})}{2}

⇥

\frac{(1 + \sqrt{13}) + (- 1 + \sqrt{13})}{2}

⇥ⓘ The numerator and the denominator are computed before the division:

1 - 1

0

■Like terms are grouped by adding their coefficients:

\sqrt{13} + \sqrt{13}

2 \cdot \sqrt{13}

⇥

\frac{0 + 2 \cdot \sqrt{13}}{2}

⇥ⓘ Adding 0 changes nothing:

0 + 2 \cdot \sqrt{13}

2 \cdot \sqrt{13}

■A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{2 \cdot \sqrt{13}}{2}

\sqrt{13}

⇥

\sqrt{13}

So the area is

\frac{2 d^{3}}{6}

⇥

\frac{2 \cdot {(\sqrt{13})}^{3}}{6}

⇥ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{2 \cdot {(\sqrt{13})}^{3}}{6}

\frac{{(\sqrt{13})}^{3}}{3}

⇥

\frac{{(\sqrt{13})}^{3}}{3}

⇥ⓘ A power of a power multiplies the exponents, (a^n)^m = a^(n*m):

{(\sqrt{13})}^{3}

13^{\frac{1}{2} \cdot 3}

⇥

\frac{13^{\frac{1}{2} \cdot 3}}{3}

⇥ⓘ Multiplying a fraction by a number multiplies its numerator:

\frac{1}{2} \cdot 3

\frac{1 \cdot 3}{2}

\frac{3}{2}

⇥

\frac{13^{\frac{3}{2}}}{3}

⇥ⓘ Powers are computed before multiplications, additions and subtractions:

13^{\frac{3}{2}}

13 \cdot \sqrt{13}

⇥

\frac{13 \cdot \sqrt{13}}{3}

The area on

[\frac{1 - \sqrt{13}}{2}, \frac{1 + \sqrt{13}}{2}]

\frac{13 \cdot \sqrt{13}}{3}

▶Area:

\frac{13 \cdot \sqrt{13}}{3}

2) The numeric value of:

\frac{13 \cdot \sqrt{13}}{3}

is:

\frac{13 \cdot \approx 3.6056}{3}

\frac{\approx 46.8722}{3}

▷

\approx 15.6241