area y=12-x^2,y=x^2-6 — step-by-step solution

1) Consider the two curves with equations

y = 12 - x^{2}

and

y = x^{2} - 6

At the intersections,

12 - x^{2}

x^{2} - 6

. Subtract the second expression from the first:

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

= 0, giving

18 - 2 x^{2} = 0

⇥1.1) Let's solve in

ℝ

18 - 2 x^{2} = 0

⇥ⓘ Both sides are divided by

-2

⇥

x^{2} = 9

⇥The two numbers whose square is

9

are

-3

and

3

⇥Solutions found:

x = {-3, 3}

The area between

12 - x^{2}

and

x^{2} - 6

[-3, 3]

is the integral of the absolute value of their difference

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

[-3, 3]

12 - x^{2}

is above

x^{2} - 6

: integrate

18 - 2 x^{2}

⇥1.2) Let's compute the integral:

\int_{-3}^{3} (18 - 2 x^{2}) d x

⇥ⅰ

\int (18 - 2 x^{2}) d x

\int 18 d x - \int 2 x^{2} d x

⇥

\int 18 d x - \int 2 x^{2} d x

⇥ⓘ The derivative of

18 x

18

, so

18 x

is an antiderivative of

18

\int 18 d x

18 x

■A constant factor stays in front of the integral. An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

- \int 2 x^{2} d x

- 2 \cdot \frac{x^{3}}{3}

⇥

18 x - 2 \cdot \frac{x^{3}}{3}

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

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

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

▷

18 x - \frac{2 x^{3}}{3}

⇥1.3) Let's evaluate over the interval

[-3, 3]

(18 \cdot 3 - \frac{2 \cdot 3^{3}}{3}) - (18 \cdot (-3) - \frac{2 \cdot {(-3)}^{3}}{3})

⇥ⓘ Multiplications are done before additions and subtractions:

18 \cdot 3

(10 + 8) \cdot 3

10 \cdot 3 + 8 \cdot 3

30 + 24

54

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

3^{3}

3 \cdot 3 \cdot 3

27

■A minus sign before a parenthesis changes the sign of each of its terms:

- (18 \cdot (-3) - \frac{2 \cdot {(-3)}^{3}}{3})

- 18 \cdot (-3) + \frac{2 \cdot {(-3)}^{3}}{3}

⇥

54

- \frac{2 \cdot 27}{3}

- 18 \cdot (-3)

+ \frac{2 \cdot {(-3)}^{3}}{3}

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

-18 \cdot (-3)

18 \cdot 3

(10 + 8) \cdot 3

10 \cdot 3 + 8 \cdot 3

30 + 24

54

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

{(-3)}^{3}

(-3) \cdot (-3) \cdot (-3)

-27

⇥ⅰ

2 \cdot 27

2 \cdot (20 + 7)

2 \cdot 20 + 2 \cdot 7

40 + 14

54

⇥

54

- \frac{54}{3}

+ 54

+ \frac{2 \cdot (-27)}{3}

⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

3

, their greatest common divisor:

- \frac{54}{3}

-18

■A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

2 \cdot (-27)

- 2 \cdot 27

- 2 \cdot (20 + 7)

- (2 \cdot 20 + 2 \cdot 7)

- (40 + 14)

-54

⇥

54

- 18

+ 54

- \frac{54}{3}

⇥ⓘ Start from

54

and move back

18

on the number line:

54 - 18

36

■We simplify the fraction by dividing the numerator and the denominator by

3

, their greatest common divisor:

- \frac{54}{3}

-18

⇥

36 + 54 - 18

⇥ⓘ Start from

36

and move forward

54

on the number line:

36 + 54

90

⇥

90 - 18

⇥ⓘ Start from

90

and move back

18

on the number line:

90 - 18

72

⇥

72

The area on

[-3, 3]

72

▶Area:

72