What is the integral from 0 to 4 of sqrt(1+\sqrt{x)} ?

The integral from 0 to 4 of sqrt(1+\sqrt{x)} is 2(2/3+(18sqrt(3)-2)/5-2sqrt(3))

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integral from 0 to 4 of sqrt(1+\sqrt{x)}

\int_{0}^{4} 1 + x d x

2 (\frac{2}{3} + \frac{18 3 - 2}{5} - 23)

Decimal

6.07589 \dots

Solution steps

\int_{0}^{4} 1 + x d x

Apply u-substitution

: \int_{1}^{3} u \cdot 2 (u - 1) d u

= \int_{1}^{3} 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}^{3} u (u - 1) d u

Expand

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

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

\int_{1}^{3} u^{\frac{3}{2}} d u = \frac{18 3 - 2}{5}

\int_{1}^{3} u d u = 23 - \frac{2}{3}

= 2 (\frac{18 3 - 2}{5} - (23 - \frac{2}{3}))

Simplify

= 2 (\frac{2}{3} + \frac{18 3 - 2}{5} - 23)

\frac{\partial}{\partial x} (2^{x} + 2^{y})

\frac{\partial}{\partial x} (3 x^{2} - 6 x + 3 y^{2})

\int x^{2} e d x

sin (100 \cdot t)

8 y^{^{''}} - 8 y^{^{'}} + 52 y = 0

What is the integral from 0 to 4 of sqrt(1+\sqrt{x)} ?
The integral from 0 to 4 of sqrt(1+\sqrt{x)} is 2(2/3+(18sqrt(3)-2)/5-2sqrt(3))