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

The integral from 0 to 1 of 6(1+sqrt(x))^7 is 1793/6

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integral from 0 to 1 of 6(1+sqrt(x))^7

\int_{0}^{1} 6 (1 + x)^{7} d x

\frac{1793}{6}

Decimal Notation

298.83333 \dots

Solution steps

\int_{0}^{1} 6 (1 + x)^{7} 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)^{7} d x

Expand

(1 + x)^{7} : 1 + 7 x + 21 x + 35 x^{\frac{3}{2}} + 35 x^{2} + 21 x^{\frac{5}{2}} + 7 x^{3} + x^{\frac{7}{2}}

= 6 \cdot \int_{0}^{1} 1 + 7 x + 21 x + 35 x^{\frac{3}{2}} + 35 x^{2} + 21 x^{\frac{5}{2}} + 7 x^{3} + x^{\frac{7}{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} 7 x d x + \int_{0}^{1} 21 x d x + \int_{0}^{1} 35 x^{\frac{3}{2}} d x + \int_{0}^{1} 35 x^{2} d x + \int_{0}^{1} 21 x^{\frac{5}{2}} d x + \int_{0}^{1} 7 x^{3} d x + \int_{0}^{1} x^{\frac{7}{2}} d x)

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

\int_{0}^{1} 7 x d x = \frac{14}{3}

\int_{0}^{1} 21 x d x = \frac{21}{2}

\int_{0}^{1} 35 x^{\frac{3}{2}} d x = 14

\int_{0}^{1} 35 x^{2} d x = \frac{35}{3}

\int_{0}^{1} 21 x^{\frac{5}{2}} d x = 6

\int_{0}^{1} 7 x^{3} d x = \frac{7}{4}

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

= 6 (1 + \frac{14}{3} + \frac{21}{2} + 14 + \frac{35}{3} + 6 + \frac{7}{4} + \frac{2}{9})

Simplify

6 (1 + \frac{14}{3} + \frac{21}{2} + 14 + \frac{35}{3} + 6 + \frac{7}{4} + \frac{2}{9}) : \frac{1793}{6}

= \frac{1793}{6}

f (x) = (2 - x^{3})^{2}

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

(x, y) \to (0, 0) lim (x^{y})

x \to 2 lim (7 - 2 x)

\int \frac{x}{x ^{4} + 9} d x

What is the integral from 0 to 1 of 6(1+sqrt(x))^7 ?
The integral from 0 to 1 of 6(1+sqrt(x))^7 is 1793/6