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

The integral from 0 to 2 of (2x+1)^3 is 78

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

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

78

Solution steps

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

Apply u-substitution

: \int_{1}^{5} \frac{u ^{3}}{2} d u

= \int_{1}^{5} \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}^{5} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{1}^{5}

= \frac{1}{2} [\frac{u ^{4}}{4}]_{1}^{5}

Compute the boundaries:

156

= \frac{1}{2} \cdot 156

Simplify

= 78

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

\int_{0}^{3} 4 x e^{2 x^{2}} d x

\int_{0}^{3} \frac{x}{25 - x ^{2}} d x

\int_{0}^{4} x + 5 d x

\int_{1}^{2} 1 - \frac{x}{2} d x

What is the integral from 0 to 2 of (2x+1)^3 ?
The integral from 0 to 2 of (2x+1)^3 is 78