area y=x^2-1,y=-x^2+1 — step-by-step solution

1) Consider the two curves with equations

y = x^{2} - 1

and

y = 1 - x^{2}

At the intersections,

x^{2} - 1

1 - x^{2}

. Subtract the second expression from the first:

(x^{2} - 1) - (1 - x^{2})

= 0, giving

2 x^{2} - 2 = 0

⇥1.1) Let's solve in

ℝ

2 x^{2} - 2 = 0

⇥ⓘ Both sides are divided by

2

⇥

x^{2} = 1

⇥The two numbers whose square is

1

are

-1

and

1

⇥Solutions found:

x = {-1, 1}

The area between

x^{2} - 1

and

1 - x^{2}

[-1, 1]

is the integral of the absolute value of their difference

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

[-1, 1]

1 - x^{2}

is above

x^{2} - 1

: integrate

- 2 x^{2} + 2

⇥1.2) Let's compute the integral:

\int_{-1}^{1} (- 2 x^{2} + 2) d x

⇥ⅰ

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

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

⇥

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

■The derivative of

2 x

2

, so

2 x

is an antiderivative of

2

\int 2 d x

2 x

⇥

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

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

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

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

▷

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

⇥1.3) Let's evaluate over the interval

[-1, 1]

(2 \cdot 1 - \frac{2 \cdot 1^{3}}{3}) - (2 \cdot (-1) - \frac{2 \cdot {(-1)}^{3}}{3})

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

- (2 \cdot (-1) - \frac{2 \cdot {(-1)}^{3}}{3})

- 2 \cdot (-1) + \frac{2 \cdot {(-1)}^{3}}{3}

⇥

2

- \frac{2}{3}

- 2 \cdot (-1)

+ \frac{2 \cdot {(-1)}^{3}}{3}

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

2 - \frac{2}{3}

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

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

-2 \cdot (-1)

2

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

{(-1)}^{3}

(-1) \cdot (-1) \cdot (-1)

-1

■Multiplying by -1 gives the opposite:

2 \cdot (-1)

-2

⇥

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

⇥ⓘ Multiplications are done before additions and subtractions:

3 \cdot 2

6

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

2 - \frac{2}{3}

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

⇥

\frac{6 - 2}{3} + \frac{3 \cdot 2 - 2}{3}

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

6 - 2

4

■Multiplications are done before additions and subtractions:

3 \cdot 2

6

⇥

\frac{4}{3} + \frac{6 - 2}{3}

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

6 - 2

4

⇥

\frac{4}{3} + \frac{4}{3}

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

\frac{4}{3} + \frac{4}{3}

\frac{4 + 4}{3}

⇥

\frac{4 + 4}{3}

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

4 + 4

8

⇥

\frac{8}{3}

The area on

[-1, 1]

\frac{8}{3}

▶Area:

\frac{8}{3}

2) The numeric value of:

\frac{8}{3}

equals:

▷

\approx 2.6667