What is the integral from 0 to 1 of (1+sqrt(x))^8 ?

The integral from 0 to 1 of (1+sqrt(x))^8 is 4097/45

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integral from 0 to 1 of (1+sqrt(x))^8

\int_{0}^{1} (1 + x)^{8} d x

\frac{4097}{45}

Decimal Notation

91.04444 \dots

Solution steps

\int_{0}^{1} (1 + x)^{8} d x

Expand

(1 + x)^{8} : 1 + 8 x + 28 x + 56 x^{\frac{3}{2}} + 70 x^{2} + 56 x^{\frac{5}{2}} + 28 x^{3} + 8 x^{\frac{7}{2}} + x^{4}

= \int_{0}^{1} 1 + 8 x + 28 x + 56 x^{\frac{3}{2}} + 70 x^{2} + 56 x^{\frac{5}{2}} + 28 x^{3} + 8 x^{\frac{7}{2}} + x^{4} 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} 1 d x + \int_{0}^{1} 8 x d x + \int_{0}^{1} 28 x d x + \int_{0}^{1} 56 x^{\frac{3}{2}} d x + \int_{0}^{1} 70 x^{2} d x + \int_{0}^{1} 56 x^{\frac{5}{2}} d x + \int_{0}^{1} 28 x^{3} d x + \int_{0}^{1} 8 x^{\frac{7}{2}} d x + \int_{0}^{1} x^{4} d x

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

\int_{0}^{1} 8 x d x = \frac{16}{3}

\int_{0}^{1} 28 x d x = 14

\int_{0}^{1} 56 x^{\frac{3}{2}} d x = \frac{112}{5}

\int_{0}^{1} 70 x^{2} d x = \frac{70}{3}

\int_{0}^{1} 56 x^{\frac{5}{2}} d x = 16

\int_{0}^{1} 28 x^{3} d x = 7

\int_{0}^{1} 8 x^{\frac{7}{2}} d x = \frac{16}{9}

\int_{0}^{1} x^{4} d x = \frac{1}{5}

= 1 + \frac{16}{3} + 14 + \frac{112}{5} + \frac{70}{3} + 16 + 7 + \frac{16}{9} + \frac{1}{5}

Simplify

1 + \frac{16}{3} + 14 + \frac{112}{5} + \frac{70}{3} + 16 + 7 + \frac{16}{9} + \frac{1}{5} : \frac{4097}{45}

= \frac{4097}{45}

\frac{\partial}{\partial x} (e^{3 x yz - 2})

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

x \to \frac{π}{2} lim (\frac{sin ( x )}{x})

x = \frac{y ^{2}}{4}, x = \frac{y}{2} + 2

n = 0 \sum \infty n sin (nπ)

What is the integral from 0 to 1 of (1+sqrt(x))^8 ?
The integral from 0 to 1 of (1+sqrt(x))^8 is 4097/45