What is the integral from 0 to 1 of sqrt(1+\sqrt{x)} ?

The integral from 0 to 1 of sqrt(1+\sqrt{x)} is 2*(4sqrt(2)+4)/(15)

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integral from 0 to 1 of sqrt(1+\sqrt{x)}

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

2 \cdot \frac{4 2 + 4}{15}

Decimal

1.28758 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{2} u \cdot 2 (u - 1) d u

= \int_{1}^{2} u \cdot 2 (u - 1) d u

Take the constant out:

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

= 2 \cdot \int_{1}^{2} u (u - 1) d u

Expand

u (u - 1) : u^{\frac{3}{2}} - u

= 2 \cdot \int_{1}^{2} u^{\frac{3}{2}} - u d u

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_{1}^{2} u^{\frac{3}{2}} d u - \int_{1}^{2} u d u)

\int_{1}^{2} u^{\frac{3}{2}} d u = \frac{8 2 - 2}{5}

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

= 2 (\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3})

Join

\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3} : \frac{4 2 + 4}{15}

= 2 \cdot \frac{4 2 + 4}{15}

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

(\frac{6}{x})^{^{'}}

\frac{d y}{d x} - y = 2 e^{x}

x \to \infty lim ((5)^{x})

\frac{\partial}{\partial x} (e^{s i n (x)} cos (x))

What is the integral from 0 to 1 of sqrt(1+\sqrt{x)} ?
The integral from 0 to 1 of sqrt(1+\sqrt{x)} is 2*(4sqrt(2)+4)/(15)