What is the integral from 0 to 1 of (2x+1)^4 ?

The integral from 0 to 1 of (2x+1)^4 is 121/5

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integral from 0 to 1 of (2x+1)^4

\int_{0}^{1} (2 x + 1)^{4} d x

\frac{121}{5}

Decimal Notation

24.2

Solution steps

\int_{0}^{1} (2 x + 1)^{4} d x

Apply u-substitution

: \int_{1}^{3} \frac{u ^{4}}{2} d u

= \int_{1}^{3} \frac{u ^{4}}{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}^{3} u^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{1}^{3}

= \frac{1}{2} [\frac{u ^{5}}{5}]_{1}^{3}

Compute the boundaries:

\frac{242}{5}

= \frac{1}{2} \cdot \frac{242}{5}

Simplify

= \frac{121}{5}

f (x) = - 8 sin (x)

\int (z^{2}) d z

\frac{\partial}{\partial x} (\frac{1}{x} + \frac{1}{1 - x})

\int (- 2 cos (x)) d x

\int sin (\frac{t}{2}) t d t

What is the integral from 0 to 1 of (2x+1)^4 ?
The integral from 0 to 1 of (2x+1)^4 is 121/5