What is the integral from 0 to 1 of (3+xsqrt(x)) ?

The integral from 0 to 1 of (3+xsqrt(x)) is 17/5

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integral from 0 to 1 of (3+xsqrt(x))

\int_{0}^{1} (3 + x x) d x

\frac{17}{5}

Decimal

3.4

Solution steps

\int_{0}^{1} 3 + x x 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

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

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

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

= 3 + \frac{2}{5}

Simplify

3 + \frac{2}{5} : \frac{17}{5}

= \frac{17}{5}

\int ln (x^{5}) d x

\frac{d}{d x} (ln (8 x^{2}))

x^{2} + 4 x - 2

f (t) = e^{2 t} cos (3 t)

n = 1 \sum \infty 2^{- n} + 3^{- n}

What is the integral from 0 to 1 of (3+xsqrt(x)) ?
The integral from 0 to 1 of (3+xsqrt(x)) is 17/5