area x,0,-1,1 — step-by-step solution

1) The area between

x

and

0

[-1, 1]

is the integral of the absolute value of their difference

Split the interval at the interior intersections:

(-1, 0, 1)

[-1, 0]

0

is above

x

: integrate

- x

⇥1.1) Let's integrate:

\int_{-1}^{0} (- x) d x

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

\int (- x) d x

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

▷

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

⇥1.2) Let's evaluate over the interval

[-1, 0]

- \frac{0^{2}}{2} + \frac{{(-1)}^{2}}{2}

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

{(-1)}^{2}

(-1) \cdot (-1)

1

⇥

0 + \frac{1}{2}

⇥ⓘ Adding 0 changes nothing:

0 + \frac{1}{2}

\frac{1}{2}

⇥

\frac{1}{2}

The area on

[-1, 0]

\frac{1}{2}

[0, 1]

x

is above

0

: integrate

x

⇥1.3) Let's compute the integral:

\int_{0}^{1} x d x

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

\int x d x

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

▷

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

⇥1.4) Let's evaluate for the interval

[0, 1]

\frac{1^{2}}{2} - \frac{0^{2}}{2}

⇥

\frac{1}{2} + 0

⇥ⓘ Adding 0 changes nothing:

\frac{1}{2} + 0

\frac{1}{2}

⇥

\frac{1}{2}

The area on

[0, 1]

\frac{1}{2}

⇥

\frac{1}{2} + \frac{1}{2}

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

\frac{1}{2} + \frac{1}{2}

\frac{1 + 1}{2}

⇥

\frac{1 + 1}{2}

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

1 + 1

2

⇥

\frac{2}{2}

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

2

, their greatest common divisor:

\frac{2}{2}

1

▶Area:

1