integral from 0 to 1 of (1-x^2)^2 — step-by-step solution

1) Let's compute the integral:

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

⇥For

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

, we use integration by parts:

\int u1 \cdot v2' d x = u1 \cdot v2 - \int u1' \cdot v2 d x

, with u1 =

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

and v2' =

1

⇥1.1) Let's differentiate u1:

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

⇥Apply the differentiation rule:

\frac{d}{d x} (f^{n}) = n \cdot \frac{d}{d x} (f) \cdot f^{n - 1}

with f =

1 - x^{2}

⇥ⓘ The derivative of x^n is n*x^(n-1):

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

2 \cdot \frac{d}{d x} (1 - x^{2}) \cdot (1 - x^{2})

⇥

2 \cdot \frac{d}{d x} (1 - x^{2}) \cdot (1 - x^{2})

⇥ⓘ Expand by multiplying each term inside the parentheses:

2 \cdot \frac{d}{d x} (1 - x^{2}) \cdot (1 - x^{2})

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

■The derivative of a sum is the sum of the derivatives:

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

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

⇥

2 \cdot (0 + \frac{d}{d x} (- x^{2})) + 2 \cdot (0 + \frac{d}{d x} (- x^{2})) \cdot (- x^{2})

⇥ⓘ Adding 0 changes nothing:

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

\frac{d}{d x} (- x^{2})

■The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (- x^{2})

- 2 x

■The minus sign is placed in front of the product:

2 \cdot (- x^{2})

-2 x^{2}

⇥

2 \cdot (- 2 x) - 2 \cdot \frac{d}{d x} (- x^{2}) \cdot x^{2}

⇥ⓘ Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

2 \cdot (-2)

-4

■The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (- x^{2})

- 2 x

⇥

- 4 x - 2 \cdot (- 2 x) \cdot x^{2}

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

-2 \cdot (-2)

4

■Multiplying powers with the same base adds the exponents:

x x^{2}

x^{3}

⇥

- 4 x + 4 x^{3}

⇥So

u1'

is equal to

- 4 x + 4 x^{3}

⇥1.2) Let's compute the integral v2':

\int 1 d x

⇥ⓘ The derivative of

x

1

, so

x

is an antiderivative of

1

\int 1 d x

x

⇥

x

⇥1.3) Let's evaluate

u1' \cdot v2

(- 4 x + 4 x^{3}) \cdot x

⇥ⓘ Expand by multiplying each term inside the parentheses:

(- 4 x + 4 x^{3}) \cdot x

- 4 x^{2} + 4 x^{4}

⇥

- 4 x^{2} + 4 x^{4}

⇥1.4) Let's calculate

u1 \cdot v2 - \int u1' \cdot v2 d x

{(1 - x^{2})}^{2} \cdot x - \int (- 4 x^{2} + 4 x^{4}) d x

⇥ⅰ

- \int (- 4 x^{2} + 4 x^{4}) d x

- (- \int 4 x^{2} d x + \int 4 x^{4} d x)

⇥

{(1 - x^{2})}^{2} \cdot x - (- \int 4 x^{2} d x + \int 4 x^{4} d x)

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

- (- \int 4 x^{2} d x + \int 4 x^{4} d x)

\int 4 x^{2} d x - \int 4 x^{4} d x

⇥

{(1 - x^{2})}^{2} \cdot x + \int 4 x^{2} d x - \int 4 x^{4} 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 4 x^{2} d x

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

- \int 4 x^{4} d x

- 4 \cdot \frac{x^{5}}{5}

⇥

{(1 - x^{2})}^{2} \cdot x + 4 \cdot \frac{x^{3}}{3} - 4 \cdot \frac{x^{5}}{5}

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

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

\frac{4 x^{3}}{3}

4 \cdot \frac{x^{5}}{5}

\frac{4 x^{5}}{5}

⇥

{(1 - x^{2})}^{2} \cdot x + \frac{4 x^{3}}{3} - \frac{4 x^{5}}{5}

You can then resume the integral computation:

ⅰ

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

{(1 - x^{2})}^{2} \cdot x + \frac{4 x^{3}}{3} - \frac{4 x^{5}}{5}

▷

{(1 - x^{2})}^{2} \cdot x + \frac{4 x^{3}}{3} - \frac{4 x^{5}}{5}

2) Let's evaluate for the interval

[0, 1]

(- \frac{4 \cdot 1^{5}}{5} + \frac{4 \cdot 1^{3}}{3} + 1 \cdot {(1 - 1^{2})}^{2}) - (- \frac{4 \cdot 0^{5}}{5} + \frac{4 \cdot 0^{3}}{3} + 0 \cdot {(1 - 0^{2})}^{2})

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

- (- \frac{4 \cdot 0^{5}}{5} + \frac{4 \cdot 0^{3}}{3} + 0 \cdot {(1 - 0^{2})}^{2})

\frac{4 \cdot 0^{5}}{5} - \frac{4 \cdot 0^{3}}{3} + 0 \cdot {(1 - 0^{2})}^{2}

- \frac{4}{5}

+ \frac{4}{3}

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

+ \frac{4 \cdot 0^{5}}{5}

- \frac{4 \cdot 0^{3}}{3}

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

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

15

, the least common multiple of

5

and

3

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

- \frac{4}{5} + \frac{4}{3}

\frac{-4 \cdot 3}{5 \cdot 3} + \frac{4 \cdot 5}{3 \cdot 5}

\frac{- 4 \cdot 3 + 4 \cdot 5}{5 \cdot 3}

■Parentheses are computed first:

1 - 1

0

■A product with a zero factor is zero:

4 \cdot 0

0

0 \cdot 1

0

■Adding 0 changes nothing:

1 + 0

1

\frac{- 4 \cdot 3 + 4 \cdot 5}{5 \cdot 3}

+ 0^{2}

+ 0

+ 0

+ 0

ⓘ Adding 0 changes nothing:

\frac{- 4 \cdot 3 + 4 \cdot 5}{5 \cdot 3}

+ 0

+ 0

+ 0

+ 0

\frac{- 4 \cdot 3 + 4 \cdot 5}{5 \cdot 3}

■Multiplications are done before additions and subtractions:

4 \cdot 3

12

4 \cdot 5

20

ⅰ

5 \cdot 3

15

\frac{- 12 + 20}{15}

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

- 12 + 20

8

▶Integral:

\frac{8}{15}

3) The numeric value of:

\frac{8}{15}

is:

▷

\approx 0.5333