What is the integral from 0 to 4 of 2pix(8-x^{3/2}) ?

The integral from 0 to 4 of 2pix(8-x^{3/2}) is (384pi)/7

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integral from 0 to 4 of 2pix(8-x^{3/2})

\int_{0}^{4} 2 π x (8 - x^{\frac{3}{2}}) d x

\frac{384 π}{7}

Decimal

172.33879 \dots

Solution steps

\int_{0}^{4} 2 π x (8 - x^{\frac{3}{2}}) d x

Take the constant out:

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

= 2 π \cdot \int_{0}^{4} x (8 - x^{\frac{3}{2}}) d x

Expand

x (8 - x^{\frac{3}{2}}) : 8 x - x^{\frac{5}{2}}

= 2 π \cdot \int_{0}^{4} 8 x - x^{\frac{5}{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 π (\int_{0}^{4} 8 x d x - \int_{0}^{4} x^{\frac{5}{2}} d x)

\int_{0}^{4} 8 x d x = 64

\int_{0}^{4} x^{\frac{5}{2}} d x = \frac{256}{7}

= 2 π (64 - \frac{256}{7})

Simplify

2 π (64 - \frac{256}{7}) : \frac{384 π}{7}

= \frac{384 π}{7}

\int_{2}^{3} \frac{33}{3 - x} d x

cos^{2} (16 t)

\int (x + 1) sin (x^{2} + 2 x + 3) d x

\frac{d x}{d t} = e^{x} \cdot sin (t)

sin (\frac{x π}{4})

What is the integral from 0 to 4 of 2pix(8-x^{3/2}) ?
The integral from 0 to 4 of 2pix(8-x^{3/2}) is (384pi)/7