What is the integral from-1 to 0 of-5x(3x^2-2)^3 ?

The integral from-1 to 0 of-5x(3x^2-2)^3 is -25/8

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integral from-1 to 0 of-5x(3x^2-2)^3

\int_{- 1}^{0} - 5 x (3 x^{2} - 2)^{3} d x

- \frac{25}{8}

Decimal Notation

- 3.125

Solution steps

\int_{- 1}^{0} - 5 x (3 x^{2} - 2)^{3} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 5 \cdot \int_{- 1}^{0} x (3 x^{2} - 2)^{3} d x

Apply u-substitution

: \int_{1}^{- 2} \frac{u ^{3}}{6} d u

= - 5 \cdot \int_{1}^{- 2} \frac{u ^{3}}{6} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - 5 (- \int_{- 2}^{1} \frac{u ^{3}}{6} d u)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 5 (- \frac{1}{6} \cdot \int_{- 2}^{1} u^{3} d u)

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{- 2}^{1}

= - 5 (- \frac{1}{6} [\frac{u ^{4}}{4}]_{- 2}^{1})

Simplify

- 5 (- \frac{1}{6} [\frac{u ^{4}}{4}]_{- 2}^{1}) : \frac{5}{6} [\frac{u ^{4}}{4}]_{- 2}^{1}

= \frac{5}{6} [\frac{u ^{4}}{4}]_{- 2}^{1}

Compute the boundaries:

- \frac{15}{4}

= \frac{5}{6} (- \frac{15}{4})

Simplify

= - \frac{25}{8}

\frac{d P}{d t} = P (1 0^{- 1} - 1 0^{- 7} P), P (0) = 7000

\int (25 - x^{2}) d x

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

\frac{d}{d x} (\frac{x - 1}{4})

\frac{d}{d x} (8 arcsin (x^{3}))

What is the integral from-1 to 0 of-5x(3x^2-2)^3 ?
The integral from-1 to 0 of-5x(3x^2-2)^3 is -25/8