What is the integral from 0 to 1 of 2(1+sqrt(x))^4 ?

The integral from 0 to 1 of 2(1+sqrt(x))^4 is 86/5

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integral from 0 to 1 of 2(1+sqrt(x))^4

\int_{0}^{1} 2 (1 + x)^{4} d x

\frac{86}{5}

Decimal

17.2

Solution steps

\int_{0}^{1} 2 (1 + x)^{4} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} (1 + x)^{4} d x

Expand

(1 + x)^{4} : 1 + 4 x + 6 x + 4 x^{\frac{3}{2}} + x^{2}

= 2 \cdot \int_{0}^{1} 1 + 4 x + 6 x + 4 x^{\frac{3}{2}} + x^{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 (\int_{0}^{1} 1 d x + \int_{0}^{1} 4 x d x + \int_{0}^{1} 6 x d x + \int_{0}^{1} 4 x^{\frac{3}{2}} d x + \int_{0}^{1} x^{2} d x)

\int_{0}^{1} 1 d x = 1

\int_{0}^{1} 4 x d x = \frac{8}{3}

\int_{0}^{1} 6 x d x = 3

\int_{0}^{1} 4 x^{\frac{3}{2}} d x = \frac{8}{5}

\int_{0}^{1} x^{2} d x = \frac{1}{3}

= 2 (1 + \frac{8}{3} + 3 + \frac{8}{5} + \frac{1}{3})

Simplify

2 (1 + \frac{8}{3} + 3 + \frac{8}{5} + \frac{1}{3}) : \frac{86}{5}

= \frac{86}{5}

\frac{d}{d x} (cos (x^{2}) - 2 x^{2} sin (x^{2}))

\frac{d}{d x} (1 - tan^{2} (x))

y = 1 - cos^{2} (x)

\frac{d}{d x} (- sin (\frac{x}{2}))

y^{^{'}} = 13.2 - \frac{6 y}{740 - 2 t}

What is the integral from 0 to 1 of 2(1+sqrt(x))^4 ?
The integral from 0 to 1 of 2(1+sqrt(x))^4 is 86/5