What is the integral from 0 to 1 of (2x-1)^6 ?

The integral from 0 to 1 of (2x-1)^6 is 1/7

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integral from 0 to 1 of (2x-1)^6

\int_{0}^{1} (2 x - 1)^{6} d x

\frac{1}{7}

Decimal Notation

0.14285 \dots

Solution steps

\int_{0}^{1} (2 x - 1)^{6} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: [\frac{u ^{7}}{7}]_{- 1}^{1}

= \frac{1}{2} [\frac{u ^{7}}{7}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{7}

= \frac{1}{2} \cdot \frac{2}{7}

Simplify

= \frac{1}{7}

\int \frac{1}{( 5 x - 2 ) ^{\frac{3}{2}}} d x

x \to 2 lim (sin (x - 2) + 1)

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

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

(5 x + 2) (\frac{1}{x} + 3 x^{5})

What is the integral from 0 to 1 of (2x-1)^6 ?
The integral from 0 to 1 of (2x-1)^6 is 1/7