What is the integral from 1 to 2 of (4x^{3/2}) ?

The integral from 1 to 2 of (4x^{3/2}) is (4(8*2^{1/2}-2))/5

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integral from 1 to 2 of (4x^{3/2})

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

\frac{4 ( 8 \cdot 2 ^{\frac{1}{2}} - 2 )}{5}

Decimal Notation

7.45096 \dots

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{8 \cdot 2 ^{\frac{1}{2}}}{5} - \frac{2}{5}

= 4 (\frac{8 \cdot 2 ^{\frac{1}{2}}}{5} - \frac{2}{5})

Simplify

= \frac{4 ( 8 \cdot 2 ^{\frac{1}{2}} - 2 )}{5}

\frac{\partial}{\partial y} (sin (2 x + 5 y))

\int_{1}^{4} 2 t ln (t) d t

\frac{\partial}{\partial x} (8 x^{6 y})

\frac{\partial}{\partial x} (20)

What is the integral from 1 to 2 of (4x^{3/2}) ?
The integral from 1 to 2 of (4x^{3/2}) is (4(8*2^{1/2}-2))/5