area x^2,0,-1,1 — step-by-step solution

1) The area between

x^{2}

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]

x^{2}

is above

0

: integrate

x^{2}

⇥1.1) Let's compute the integral:

\int_{-1}^{0} x^{2} 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}

▷

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

⇥1.2) Let's evaluate for the interval

[-1, 0]

\frac{0^{3}}{3} - \frac{{(-1)}^{3}}{3}

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

{(-1)}^{3}

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

-1

⇥

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

⇥ⓘ Adding 0 changes nothing:

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

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

■Two minus signs cancel out:

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

\frac{1}{3}

⇥

\frac{1}{3}

The area on

[-1, 0]

\frac{1}{3}

[0, 1]

x^{2}

is above

0

: integrate

x^{2}

⇥1.3) Let's compute the integral:

\int_{0}^{1} x^{2} 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}

▷

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

⇥1.4) Let's evaluate over the interval

[0, 1]

\frac{1^{3}}{3} - \frac{0^{3}}{3}

⇥

\frac{1}{3} + 0

⇥ⓘ Adding 0 changes nothing:

\frac{1}{3} + 0

\frac{1}{3}

⇥

\frac{1}{3}

The area on

[0, 1]

\frac{1}{3}

⇥

\frac{1}{3} + \frac{1}{3}

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

\frac{1}{3} + \frac{1}{3}

\frac{1 + 1}{3}

⇥

\frac{1 + 1}{3}

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

1 + 1

2

▶Area:

\frac{2}{3}

2) The numeric value of:

\frac{2}{3}

is:

▷

\approx 0.6667