What is the integral from 1 to 4 of sqrt(x^3) ?

The integral from 1 to 4 of sqrt(x^3) is 62/5

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integral from 1 to 4 of sqrt(x^3)

\int_{1}^{4} x^{3} d x

\frac{62}{5}

Decimal Notation

12.4

Solution steps

\int_{1}^{4} x^{3} d x

x^{3} = x^{\frac{3}{2}},

assuming

x \geq 0

= \int_{1}^{4} x^{\frac{3}{2}} d x

Apply the Power Rule

: [\frac{2}{5} x^{\frac{5}{2}}]_{1}^{4}

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

Compute the boundaries:

\frac{62}{5}

= \frac{62}{5}

\int_{3}^{9} 1 + 3 x d x

\int_{0}^{5} \frac{1}{1 + x ^{2} + y ^{2}} d z

\int_{y}^{2 y} 8 x + e^{y} d x

\int_{- 1}^{3} 6 x^{5} - 2 x^{3} + 2 d x

What is the integral from 1 to 4 of sqrt(x^3) ?
The integral from 1 to 4 of sqrt(x^3) is 62/5