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

The integral from-1 to 0 of (x+1)^3 is 1/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{4}

Decimal Notation

0.25

Solution steps

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

Apply u-substitution

: \int_{0}^{1} u^{3} d u

= \int_{0}^{1} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{0}^{1}

= [\frac{u ^{4}}{4}]_{0}^{1}

Compute the boundaries:

\frac{1}{4}

= \frac{1}{4}

\frac{d}{d x} (7 x - 4)

(y^{3} - 1) e x d x + 3 y^{2} (e x + 1) d y = 0

x \to 0 lim (4 x cot (x))

\int_{0}^{1} 8 e^{2 x} d x

What is the integral from-1 to 0 of (x+1)^3 ?
The integral from-1 to 0 of (x+1)^3 is 1/4