What is the integral from 0 to 2 of (x-1)^{25} ?

The integral from 0 to 2 of (x-1)^{25} is 0

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integral from 0 to 2 of (x-1)^{25}

\int_{0}^{2} (x - 1)^{25} d x

0

Solution steps

\int_{0}^{2} (x - 1)^{25} d x

Apply u-substitution

: \int_{- 1}^{1} u^{25} d u

= \int_{- 1}^{1} u^{25} d u

Apply the Power Rule

: [\frac{u ^{26}}{26}]_{- 1}^{1}

= [\frac{u ^{26}}{26}]_{- 1}^{1}

Compute the boundaries:

0

= 0

\int_{- \infty}^{0} (x e^{2 x}) d x

\frac{d}{d x} (cos^{4} (3 x^{2}))

n = 1 \sum \infty \frac{1}{n ^{2} + n ^{3}}

x \to 9 lim (7 x - 8)

\frac{\partial}{\partial x} (- 2 ∣ x ∣)

What is the integral from 0 to 2 of (x-1)^{25} ?
The integral from 0 to 2 of (x-1)^{25} is 0