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

1) Consider the two curves with equations

y = x^{2}

and

y = 2

At the intersections,

x^{2}

2

. Subtract the second expression from the first:

x^{2} - 2 = 0

Let's solve in

ℝ

x^{2} - 2 = 0

The two numbers whose square is

2

are

- \sqrt{2}

and

\sqrt{2}

Solutions found:

x = {- \sqrt{2}, \sqrt{2}}

The area between

x^{2}

and

2

[- \sqrt{2}, \sqrt{2}]

is the integral of the absolute value of their difference

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

[- \sqrt{2}, \sqrt{2}]

2

is above

x^{2}

: integrate

- x^{2} + 2

Between its roots

x_{1} = - \sqrt{2}

and

x_{2} = \sqrt{2}

- x^{2} + 2

-1 \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{d^{3}}{6}

Compute

d = x_{2} - x_{1}

⇥

\sqrt{2} + \sqrt{2}

⇥ⓘ Like terms are grouped by adding their coefficients:

\sqrt{2} + \sqrt{2}

2 \cdot \sqrt{2}

⇥

2 \cdot \sqrt{2}

So the area is

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

⇥

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

⇥ⓘ The power of a product is the product of the powers:

{(2 \cdot \sqrt{2})}^{3}

2^{3} \cdot {(\sqrt{2})}^{3}

⇥

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

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

2^{3}

2 \cdot 2 \cdot 2

8

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

{(\sqrt{2})}^{3}

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

⇥

\frac{8 \cdot 2^{\frac{1}{2} \cdot 3}}{6}

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

\frac{1}{2} \cdot 3

\frac{1 \cdot 3}{2}

\frac{3}{2}

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

\frac{8 \cdot 2^{\frac{3}{2}}}{6}

\frac{4 \cdot 2^{\frac{3}{2}}}{3}

⇥

\frac{4 \cdot 2^{\frac{3}{2}}}{3}

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

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

2 \cdot \sqrt{2}

⇥

\frac{4 \cdot 2 \cdot \sqrt{2}}{3}

⇥ⅰ

4 \cdot 2

8

⇥

\frac{8 \cdot \sqrt{2}}{3}

The area on

[- \sqrt{2}, \sqrt{2}]

\frac{8 \cdot \sqrt{2}}{3}

▶Area:

\frac{8 \cdot \sqrt{2}}{3}

2) The numeric value of:

\frac{8 \cdot \sqrt{2}}{3}

equals:

\frac{8 \cdot \approx 1.4142}{3}

\frac{\approx 11.3137}{3}

▷

\approx 3.7712