area between x^2 and x — step-by-step solution

1) Consider the two curves with equations

y = x^{2}

and

y = x

At the intersections,

x^{2}

x

. Subtract the second expression from the first:

x^{2} - x = 0

⇥1.1) Let's solve in

ℝ

x^{2} - x = 0

⇥You can write the expression:

x^{2} - x

in factored form:

x \cdot (x - 1)

⇥A product is zero if and only if one of its factors is zero: we solve each factor equal to 0

⇥⇥1.1.1) Let's solve in

ℝ

x = 0

⇥⇥Solution obtained:

x = 0

⇥⇥1.1.2) Let's solve in

ℝ

x - 1 = 0

⇥⇥Solution obtained:

x = 1

The area between

x^{2}

and

x

[0, 1]

is the integral of the absolute value of their difference

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

[0, 1]

x

is above

x^{2}

: integrate

- x^{2} + x

⇥1.2) Let's compute the integral:

\int_{0}^{1} (- x^{2} + x) d x

⇥ⅰ

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

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

⇥

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

\int x d x

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

▷

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

⇥1.3) Let's evaluate for the interval

[0, 1]

(\frac{1^{2}}{2} - \frac{1^{3}}{3}) - (\frac{0^{2}}{2} - \frac{0^{3}}{3})

⇥ⓘ To add fractions, put them over the same denominator, then add the numerators. We choose

6

, the least common multiple of

2

and

3

. Multiplying the numerator and the denominator by the same number does not change the fraction:

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

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

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

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

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

- (\frac{0^{2}}{2} - \frac{0^{3}}{3})

- \frac{0^{2}}{2} + \frac{0^{3}}{3}

⇥

\frac{3 - 1 \cdot 2}{2 \cdot 3} - \frac{0^{2}}{2} + \frac{0^{3}}{3}

⇥ⓘ Multiplications are done before additions and subtractions:

1 \cdot 2

2

⇥ⅰ

2 \cdot 3

6

⇥

\frac{3 - 2}{6} + 0 + 0

⇥ⓘ Adding 0 changes nothing:

\frac{3 - 2}{6} + 0 + 0

\frac{3 - 2}{6}

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

3 - 2

1

⇥

\frac{1}{6}

The area on

[0, 1]

\frac{1}{6}

▶Area:

\frac{1}{6}

2) The numeric value of:

\frac{1}{6}

is:

▷

\approx 0.1667