What is the integral from 0 to 1 of 6(1+sqrt(x))^5 ?

The integral from 0 to 1 of 6(1+sqrt(x))^5 is 642/7

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integral from 0 to 1 of 6(1+sqrt(x))^5

\int_{0}^{1} 6 (1 + x)^{5} d x

\frac{642}{7}

Decimal

91.71428 \dots

Solution steps

\int_{0}^{1} 6 (1 + x)^{5} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{0}^{1} (1 + x)^{5} d x

Expand

(1 + x)^{5} : 1 + 5 x + 10 x + 10 x^{\frac{3}{2}} + 5 x^{2} + x^{\frac{5}{2}}

= 6 \cdot \int_{0}^{1} 1 + 5 x + 10 x + 10 x^{\frac{3}{2}} + 5 x^{2} + x^{\frac{5}{2}} 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

= 6 (\int_{0}^{1} 1 d x + \int_{0}^{1} 5 x d x + \int_{0}^{1} 10 x d x + \int_{0}^{1} 10 x^{\frac{3}{2}} d x + \int_{0}^{1} 5 x^{2} d x + \int_{0}^{1} x^{\frac{5}{2}} d x)

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

\int_{0}^{1} 5 x d x = \frac{10}{3}

\int_{0}^{1} 10 x d x = 5

\int_{0}^{1} 10 x^{\frac{3}{2}} d x = 4

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

\int_{0}^{1} x^{\frac{5}{2}} d x = \frac{2}{7}

= 6 (1 + \frac{10}{3} + 5 + 4 + \frac{5}{3} + \frac{2}{7})

Simplify

6 (1 + \frac{10}{3} + 5 + 4 + \frac{5}{3} + \frac{2}{7}) : \frac{642}{7}

= \frac{642}{7}

\frac{d}{d x} ((4 x)^{2})

\frac{\partial}{\partial z} (x^{2} + 3 yz + 4 x y)

n = 0 \sum \infty 6 (0.4)^{n}

\frac{d}{d x} (k y (x)^{2})

y^{^{''}} - y = 3 e^{x}

What is the integral from 0 to 1 of 6(1+sqrt(x))^5 ?
The integral from 0 to 1 of 6(1+sqrt(x))^5 is 642/7