What is the integral from 0 to 1 of 4(1+sqrt(x))^8 ?

The integral from 0 to 1 of 4(1+sqrt(x))^8 is 16388/45

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integral from 0 to 1 of 4(1+sqrt(x))^8

\int_{0}^{1} 4 (1 + x)^{8} d x

\frac{16388}{45}

Decimal Notation

364.17777 \dots

Solution steps

\int_{0}^{1} 4 (1 + x)^{8} d x

Take the constant out:

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

= 4 \cdot \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}

= 4 \cdot \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

= 4 (\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}

= 4 (1 + \frac{16}{3} + 14 + \frac{112}{5} + \frac{70}{3} + 16 + 7 + \frac{16}{9} + \frac{1}{5})

Simplify

4 (1 + \frac{16}{3} + 14 + \frac{112}{5} + \frac{70}{3} + 16 + 7 + \frac{16}{9} + \frac{1}{5}) : \frac{16388}{45}

= \frac{16388}{45}

\int 0.6 x^{2} - 4.4 x + 60 d x

\int \frac{x + 2}{x + 3} d x

- 10 x^{2} y + y^{^{'}} = 9 x^{2}

\int \frac{1}{3 ^{x} + 1} d x

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = - 3, y^{^{'}} (0) = 5

What is the integral from 0 to 1 of 4(1+sqrt(x))^8 ?
The integral from 0 to 1 of 4(1+sqrt(x))^8 is 16388/45