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

The integral from 0 to 1 of (sqrt(x)(x+1)) is 16/15

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integral from 0 to 1 of (sqrt(x)(x+1))

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

\frac{16}{15}

Decimal Notation

1.06666 \dots

Solution steps

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

Expand

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

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

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

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

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

Simplify

\frac{2}{5} + \frac{2}{3} : \frac{16}{15}

= \frac{16}{15}

- \frac{2}{5 x}

f (x) = 3 x + 6 x, (1, 9)

\int_{5}^{\infty} e^{- \frac{x}{3}} d x

\int \frac{2 x}{9 - x ^{2}} d x

- x + tan (x)

What is the integral from 0 to 1 of (sqrt(x)(x+1)) ?
The integral from 0 to 1 of (sqrt(x)(x+1)) is 16/15