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

The integral from 0 to 1 of (2x+3)^3 is 68

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

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

68

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{3}^{5}

= \frac{1}{2} [\frac{u ^{4}}{4}]_{3}^{5}

Compute the boundaries:

136

= \frac{1}{2} \cdot 136

Simplify

= 68

\int_{0}^{\frac{π}{2}} e^{- 5 x} sin (x) d x

\int_{1}^{5} \frac{5 x}{x ^{2} + 2} d x

\int_{- 2}^{5} (10 - 5 x) d x

\int_{0}^{6} \frac{8}{729} (x + 3)^{3} d x

\int_{0}^{1} \frac{8 y}{( x ^{3} + 1 ) ^{2}} d y

What is the integral from 0 to 1 of (2x+3)^3 ?
The integral from 0 to 1 of (2x+3)^3 is 68