What is the integral from 0 to 9 of sqrt(\sqrt{x)+1} ?

The integral from 0 to 9 of sqrt(\sqrt{x)+1} is 232/15

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integral from 0 to 9 of sqrt(\sqrt{x)+1}

\int_{0}^{9} x + 1 d x

\frac{232}{15}

Decimal

15.46666 \dots

Solution steps

\int_{0}^{9} x + 1 d x

Apply u-substitution

: \int_{1}^{4} u \cdot 2 (u - 1) d u

= \int_{1}^{4} 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}^{4} u (u - 1) d u

Expand

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

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

\int_{1}^{4} u^{\frac{3}{2}} d u = \frac{62}{5}

\int_{1}^{4} u d u = \frac{14}{3}

= 2 (\frac{62}{5} - \frac{14}{3})

Simplify

2 (\frac{62}{5} - \frac{14}{3}) : \frac{232}{15}

= \frac{232}{15}

\int_{3}^{6} x^{3} + x - 2 d x

\int_{1}^{9} (\frac{4}{x} - e^{- x}) d x

\int_{1}^{e} \frac{( x ^{5} + 1 )}{x} d x

\int_{0}^{\frac{1}{2}} x^{2} e^{2 x} d x

\int_{1}^{\infty} \frac{1}{n}

What is the integral from 0 to 9 of sqrt(\sqrt{x)+1} ?
The integral from 0 to 9 of sqrt(\sqrt{x)+1} is 232/15