What is the integral from 0 to 1 of (1+2x)^3 ?

The integral from 0 to 1 of (1+2x)^3 is 10

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integral from 0 to 1 of (1+2x)^3

\int_{0}^{1} (1 + 2 x)^{3} d x

10

Solution steps

\int_{0}^{1} (1 + 2 x)^{3} d x

Apply u-substitution

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

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

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{1}^{3}

= \frac{1}{2} [\frac{u ^{4}}{4}]_{1}^{3}

Compute the boundaries:

20

= \frac{1}{2} \cdot 20

Simplify

= 10

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

\int_{0}^{1} (3 x^{2} + 4 x - \frac{1}{2}) d x

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

\int_{0}^{3} \frac{3 x - 2}{x + 1} d x

\int_{0}^{8} \frac{2 x - 4}{x + 1} d x

What is the integral from 0 to 1 of (1+2x)^3 ?
The integral from 0 to 1 of (1+2x)^3 is 10