What is the integral from-1 to 1 of (z+1)^5 ?

The integral from-1 to 1 of (z+1)^5 is 32/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 1 of (z+1)^5

\int_{- 1}^{1} (z + 1)^{5} d z

\frac{32}{3}

Decimal Notation

10.66666 \dots

Solution steps

\int_{- 1}^{1} (z + 1)^{5} d z

Apply u-substitution

: \int_{0}^{2} u^{5} d u

= \int_{0}^{2} u^{5} d u

Apply the Power Rule

: [\frac{u ^{6}}{6}]_{0}^{2}

= [\frac{u ^{6}}{6}]_{0}^{2}

Compute the boundaries:

\frac{32}{3}

= \frac{32}{3}

\int_{1}^{2} \frac{1}{x ^{2} - 6 x + 9} d x

\int_{0}^{\frac{π}{3}} x tan^{2} (x) d x

\int_{0}^{x} cos^{2} (t) d t

\int_{0}^{\frac{3 π}{4}} sin (x) d x

\int_{- 8}^{\infty} - e^{- x} d x

What is the integral from-1 to 1 of (z+1)^5 ?
The integral from-1 to 1 of (z+1)^5 is 32/3