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

The integral from 1 to 4 of sqrt(\sqrt{x)-1} is 32/15

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

\int_{1}^{4} x - 1 d x

\frac{32}{15}

Decimal Notation

2.13333 \dots

Solution steps

\int_{1}^{4} x - 1 d x

Apply u-substitution

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

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

Apply u-substitution

: \int_{0}^{1} (v + 1) v d v

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

Expand

(v + 1) v : v^{\frac{3}{2}} + v

= 2 \cdot \int_{0}^{1} v^{\frac{3}{2}} + v d v

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} v^{\frac{3}{2}} d v + \int_{0}^{1} v d v)

\int_{0}^{1} v^{\frac{3}{2}} d v = \frac{2}{5}

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

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

Simplify

2 (\frac{2}{5} + \frac{2}{3}) : \frac{32}{15}

= \frac{32}{15}

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What is the integral from 1 to 4 of sqrt(\sqrt{x)-1} ?
The integral from 1 to 4 of sqrt(\sqrt{x)-1} is 32/15