integral from 0 to 1 of (1-x^7)^2 — step-by-step solution

1) Let's integrate:

\int_{0}^{1} {(1 - x^{7})}^{2} d x

⇥For

{(1 - x^{7})}^{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^{7})}^{2}

and v2' =

1

⇥1.1) Let's perform the differentiation u1:

\frac{d}{d x} ({(1 - x^{7})}^{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^{7}

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

\frac{d}{d x} ({(1 - x^{7})}^{2})

2 \cdot \frac{d}{d x} (1 - x^{7}) \cdot (1 - x^{7})

⇥

2 \cdot \frac{d}{d x} (1 - x^{7}) \cdot (1 - x^{7})

⇥ⓘ Expand by multiplying each term inside the parentheses:

2 \cdot \frac{d}{d x} (1 - x^{7}) \cdot (1 - x^{7})

2 \cdot \frac{d}{d x} (1 - x^{7}) + 2 \cdot \frac{d}{d x} (1 - x^{7}) \cdot (- x^{7})

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

\frac{d}{d x} (1 - x^{7})

0 + \frac{d}{d x} (- x^{7})

⇥

2 \cdot (0 + \frac{d}{d x} (- x^{7})) + 2 \cdot (0 + \frac{d}{d x} (- x^{7})) \cdot (- x^{7})

⇥ⓘ Adding 0 changes nothing:

0 + \frac{d}{d x} (- x^{7})

\frac{d}{d x} (- x^{7})

■The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (- x^{7})

- 7 x^{6}

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

2 \cdot (- x^{7})

-2 x^{7}

⇥

2 \cdot (- 7 x^{6}) - 2 \cdot \frac{d}{d x} (- x^{7}) \cdot x^{7}

⇥ⓘ 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 (-7)

-14

■The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (- x^{7})

- 7 x^{6}

⇥

- 14 x^{6} - 2 \cdot (- 7 x^{6}) \cdot x^{7}

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

-2 \cdot (-7)

14

■Multiplying powers with the same base adds the exponents:

x^{6} x^{7}

x^{13}

⇥

- 14 x^{6} + 14 x^{13}

⇥Thus,

u1'

- 14 x^{6} + 14 x^{13}

⇥1.2) Let's integrate 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 calculate

u1' \cdot v2

(- 14 x^{6} + 14 x^{13}) \cdot x

⇥ⓘ Expand by multiplying each term inside the parentheses:

(- 14 x^{6} + 14 x^{13}) \cdot x

- 14 x^{7} + 14 x^{14}

⇥

- 14 x^{7} + 14 x^{14}

⇥1.4) Let's calculate

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

{(1 - x^{7})}^{2} \cdot x - \int (- 14 x^{7} + 14 x^{14}) d x

⇥ⅰ

- \int (- 14 x^{7} + 14 x^{14}) d x

- (- \int 14 x^{7} d x + \int 14 x^{14} d x)

⇥

{(1 - x^{7})}^{2} \cdot x - (- \int 14 x^{7} d x + \int 14 x^{14} d x)

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

- (- \int 14 x^{7} d x + \int 14 x^{14} d x)

\int 14 x^{7} d x - \int 14 x^{14} d x

⇥

{(1 - x^{7})}^{2} \cdot x + \int 14 x^{7} d x - \int 14 x^{14} 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 14 x^{7} d x

14 \cdot \frac{x^{8}}{8}

- \int 14 x^{14} d x

- 14 \cdot \frac{x^{15}}{15}

⇥

{(1 - x^{7})}^{2} \cdot x + 14 \cdot \frac{x^{8}}{8} - 14 \cdot \frac{x^{15}}{15}

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

14 \cdot \frac{x^{8}}{8}

\frac{14 x^{8}}{8}

14 \cdot \frac{x^{15}}{15}

\frac{14 x^{15}}{15}

■A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{14 x^{8}}{8}

\frac{7 x^{8}}{4}

⇥

{(1 - x^{7})}^{2} \cdot x + \frac{7 x^{8}}{4} - \frac{14 x^{15}}{15}

Then you can resume the integral computation:

ⅰ

\int {(1 - x^{7})}^{2} d x

{(1 - x^{7})}^{2} \cdot x + \frac{7 x^{8}}{4} - \frac{14 x^{15}}{15}

▷

{(1 - x^{7})}^{2} \cdot x + \frac{7 x^{8}}{4} - \frac{14 x^{15}}{15}

2) Let's evaluate over the interval

[0, 1]

(- \frac{14 \cdot 1^{15}}{15} + \frac{7 \cdot 1^{8}}{4} + 1 \cdot {(1 - 1^{7})}^{2}) - (- \frac{14 \cdot 0^{15}}{15} + \frac{7 \cdot 0^{8}}{4} + 0 \cdot {(1 - 0^{7})}^{2})

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

- (- \frac{14 \cdot 0^{15}}{15} + \frac{7 \cdot 0^{8}}{4} + 0 \cdot {(1 - 0^{7})}^{2})

\frac{14 \cdot 0^{15}}{15} - \frac{7 \cdot 0^{8}}{4} + 0 \cdot {(1 - 0^{7})}^{2}

- \frac{14}{15}

+ \frac{7}{4}

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

+ \frac{14 \cdot 0^{15}}{15}

- \frac{7 \cdot 0^{8}}{4}

+ 0 \cdot {(1 - 0^{7})}^{2}

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

60

, the least common multiple of

15

and

4

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

- \frac{14}{15} + \frac{7}{4}

\frac{-14 \cdot 4}{15 \cdot 4} + \frac{7 \cdot 15}{4 \cdot 15}

\frac{- 14 \cdot 4 + 7 \cdot 15}{15 \cdot 4}

■Parentheses are computed first:

1 - 1

0

■A product with a zero factor is zero:

14 \cdot 0

0

7 \cdot 0

0

0 \cdot 1

0

■Adding 0 changes nothing:

1 + 0

1

\frac{- 14 \cdot 4 + 7 \cdot 15}{15 \cdot 4}

+ 0^{2}

+ 0

+ 0

+ 0

ⓘ Adding 0 changes nothing:

\frac{- 14 \cdot 4 + 7 \cdot 15}{15 \cdot 4}

+ 0

+ 0

+ 0

+ 0

\frac{- 14 \cdot 4 + 7 \cdot 15}{15 \cdot 4}

■Multiplications are done before additions and subtractions:

14 \cdot 4

(10 + 4) \cdot 4

10 \cdot 4 + 4 \cdot 4

40 + 16

56

7 \cdot 15

7 \cdot (10 + 5)

7 \cdot 10 + 7 \cdot 5

70 + 35

105

ⅰ

15 \cdot 4

(10 + 5) \cdot 4

10 \cdot 4 + 5 \cdot 4

40 + 20

60

\frac{- 56 + 105}{60}

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

- 56 + 105

49

▶Integral:

\frac{49}{60}

3) The numeric value of:

\frac{49}{60}

equals:

▷

\approx 0.8167