integral from 0 to 1 of x^2(x^3-2)^3
1) Let's compute the integral: ∫01x2·(x3 -2)3dx
Let u = x3 -2. Then du = 3x2·dx, so x2·dx = du3
∫x2·(x3 -2)3dx = 13·∫u3du
An antiderivative of u3 is u44
Substitute u = x3 -2 back into the antiderivative
ⅰ ∫x2·(x3 -2)3dx = (x3 -2)412
▷ (x3 -2)412
2) Let's evaluate over the interval [0, 1]: (13 -2)412 -(03 -2)412
ⓘ Parentheses are computed first: 1 -2 = -1■Adding 0 changes nothing: 0 -2 = -2■Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product: (-2)4 = (-2)·(-2)·(-2)·(-2) = 16
(-1)412 -1612
ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product: (-1)4 = (-1)·(-1)·(-1)·(-1) = 1■We simplify the fraction by dividing the numerator and the denominator by 4, their greatest common divisor: -1612 = -4·43·4 = -43
112 -43
ⓘ To add fractions, put them over the same denominator, then add the numerators. We choose 12, the least common multiple of 12 and 3. Multiplying the numerator and the denominator by the same number does not change the fraction: 112 -43 = 112 + -4·43·4 = 1 -4·412
1 -4·412
ⓘ Multiplications are done before additions and subtractions: 4·4 = 16
1 -1612
ⓘ The numerator and the denominator are computed before the division: 1 -16 = -15
-1512
ⓘ We simplify the fraction by dividing the numerator and the denominator by 3, their greatest common divisor: -1512 = -5·34·3 = -54
▶Integral:   -54
3) The numeric value of: -54 is:
▷ -1.25