What is the integral from 1 to 3 of (z-2)^3 ?

The integral from 1 to 3 of (z-2)^3 is 0

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integral from 1 to 3 of (z-2)^3

\int_{1}^{3} (z - 2)^{3} d z

0

Solution steps

\int_{1}^{3} (z - 2)^{3} d z

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

0

= 0

\int_{1}^{x + c} 1 d x

\int_{- 3}^{3} 3 - ∣ x ∣ d x

\int_{0}^{4} 8 - 2 x d x

\int_{0}^{1} (1 - y) y^{n} d y

\int_{\frac{π}{8}}^{\frac{π}{4}} 3 csc^{2} (2 x) d x

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