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

The integral from-3 to-2 of 2x(3-x^2)^3 is 1295/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-3 to-2 of 2x(3-x^2)^3

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

\frac{1295}{4}

Decimal

323.75

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{- 6}^{- 1}

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

Simplify

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

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

Compute the boundaries:

- \frac{1295}{4}

= - (- \frac{1295}{4})

Simplify

= \frac{1295}{4}

\int_{- 2}^{- 1} \frac{x}{( x ^{2} + 5 ) ^{2}} d x

\int_{1}^{e} \frac{( 1 + I n x ) ^{2}}{x} d x

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

\int_{1}^{3} (x^{2} + \frac{1}{x ^{2}}) d x

\int_{- 5}^{6} (36 - x^{2}) - (6 - x) d x

What is the integral from-3 to-2 of 2x(3-x^2)^3 ?
The integral from-3 to-2 of 2x(3-x^2)^3 is 1295/4