area x^2,1,-2,2 — step-by-step solution

1) The area between

x^{2}

and

1

[-2, 2]

is the integral of the absolute value of their difference

Split the interval at the interior intersections:

(-2, -1, 1, 2)

[-2, -1]

x^{2}

is above

1

: integrate

x^{2} - 1

⇥1.1) Let's integrate:

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

⇥ⅰ

\int (x^{2} - 1) d x

\int x^{2} d x - \int 1 d x

⇥

\int x^{2} d x - \int 1 d x

⇥ⓘ An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

\int x^{2} d x

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

■The derivative of

- x

-1

, so

- x

is an antiderivative of

-1

- \int 1 d x

- x

▷

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

⇥1.2) Let's evaluate over the interval

[-2, -1]

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

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

{(-1)}^{3}

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

-1

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

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

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

⇥

- \frac{1}{3}

+ 1

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

- 2

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

- \frac{1}{3} + 1

\frac{- 1 + 3}{3}

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

{(-2)}^{3}

(-2) \cdot (-2) \cdot (-2)

-8

⇥

\frac{- 1 + 3}{3} - (\frac{-8}{3}) - 2

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

- 1 + 3

2

■Two minus signs cancel out:

- (\frac{-8}{3})

\frac{8}{3}

⇥

\frac{2}{3} + \frac{8}{3} - 2

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

\frac{2}{3} + \frac{8}{3}

\frac{2 + 8}{3}

⇥

\frac{2 + 8}{3} - 2

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

2 + 8

10

⇥

\frac{10}{3} - 2

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

\frac{10}{3} - 2

\frac{10 - 2 \cdot 3}{3}

⇥

\frac{10 - 2 \cdot 3}{3}

⇥ⓘ Multiplications are done before additions and subtractions:

2 \cdot 3

6

⇥

\frac{10 - 6}{3}

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

10 - 6

4

⇥

\frac{4}{3}

The area on

[-2, -1]

\frac{4}{3}

[-1, 1]

1

is above

x^{2}

: integrate

- x^{2} + 1

⇥1.3) Let's integrate:

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

⇥ⅰ

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

- \int x^{2} d x + \int 1 d x

⇥

- \int x^{2} d x + \int 1 d x

⇥ⓘ An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

- \int x^{2} d x

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

■The derivative of

x

1

, so

x

is an antiderivative of

1

\int 1 d x

x

▷

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

⇥1.4) Let's evaluate for the interval

[-1, 1]

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

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

1 - \frac{1}{3}

\frac{3 - 1}{3}

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

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

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

⇥

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

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

3 - 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

⇥

\frac{2}{3} + 1 - \frac{1}{3}

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

\frac{2}{3} + 1

\frac{2 + 3}{3}

⇥

\frac{2 + 3}{3} - \frac{1}{3}

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

2 + 3

5

⇥

\frac{5}{3} - \frac{1}{3}

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

\frac{5}{3} - \frac{1}{3}

\frac{5 - 1}{3}

⇥

\frac{5 - 1}{3}

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

5 - 1

4

⇥

\frac{4}{3}

The area on

[-1, 1]

\frac{4}{3}

[1, 2]

x^{2}

is above

1

: integrate

x^{2} - 1

⇥1.5) Let's integrate:

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

⇥ⅰ

\int (x^{2} - 1) d x

\int x^{2} d x - \int 1 d x

⇥

\int x^{2} d x - \int 1 d x

⇥ⓘ An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

\int x^{2} d x

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

■The derivative of

- x

-1

, so

- x

is an antiderivative of

-1

- \int 1 d x

- x

▷

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

⇥1.6) Let's evaluate for the interval

[1, 2]

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

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

2^{3}

2 \cdot 2 \cdot 2

8

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

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

- \frac{1^{3}}{3} + 1

⇥

\frac{8}{3}

- 2

- \frac{1^{3}}{3}

+ 1

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

\frac{8}{3} - 2

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

⇥

\frac{8 - 2 \cdot 3}{3} - \frac{1}{3} + 1

⇥ⓘ Multiplications are done before additions and subtractions:

2 \cdot 3

6

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

- \frac{1}{3} + 1

\frac{- 1 + 3}{3}

⇥

\frac{8 - 6}{3} + \frac{- 1 + 3}{3}

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

8 - 6

2

- 1 + 3

2

⇥

\frac{2}{3} + \frac{2}{3}

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

\frac{2}{3} + \frac{2}{3}

\frac{2 + 2}{3}

⇥

\frac{2 + 2}{3}

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

2 + 2

4

⇥

\frac{4}{3}

The area on

[1, 2]

\frac{4}{3}

⇥

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

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

4 + 4

8

⇥

\frac{8}{3} + \frac{4}{3}

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

\frac{8}{3} + \frac{4}{3}

\frac{8 + 4}{3}

⇥

\frac{8 + 4}{3}

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

8 + 4

12

⇥

\frac{12}{3}

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

3

, their greatest common divisor:

\frac{12}{3}

4

▶Area:

4