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

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

Popular >

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

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

\frac{16}{15}

Decimal Notation

1.06666 \dots

Solution steps

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

Expand

(x + 1) x : 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}

\int_{0}^{\frac{π}{6}} (cos (x)) d x

\int_{- 2}^{0} (3 x - 2 e^{x}) d x

\int_{0}^{2} - x^{3} + x^{2} + 2 x d x

\int_{0}^{8} ∣ x - 3 ∣ d x

\int_{0}^{1} 4 x - 6 (x^{2})^{\frac{1}{3}} d x

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