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

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

= 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}

\int cos (3 x) cos (x) d x

3 (1 + x^{2}) y^{^{'}} = 2 x y (y^{3} - 1)

f (t) = 2 - \frac{4}{t}, (4, 1)

\frac{d}{d x} (x^{2} + \frac{a ^{4}}{x ^{2}})

\frac{d y}{d x} = \frac{x - 1}{y - 4}

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