What is the integral from 0 to 4 of (3x+4)^{3/2} ?

The integral from 0 to 4 of (3x+4)^{3/2} is 1984/15

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integral from 0 to 4 of (3x+4)^{3/2}

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

\frac{1984}{15}

Decimal Notation

132.26666 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{16} \frac{u ^{\frac{3}{2}}}{3} d u

= \int_{4}^{16} \frac{u ^{\frac{3}{2}}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{4}^{16} u^{\frac{3}{2}} d u

Apply the Power Rule

: [\frac{2}{5} u^{\frac{5}{2}}]_{4}^{16}

= \frac{1}{3} [\frac{2}{5} u^{\frac{5}{2}}]_{4}^{16}

Compute the boundaries:

\frac{1984}{5}

= \frac{1}{3} \cdot \frac{1984}{5}

Simplify

= \frac{1984}{15}

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What is the integral from 0 to 4 of (3x+4)^{3/2} ?
The integral from 0 to 4 of (3x+4)^{3/2} is 1984/15