What is the integral from 0 to 1 of 3(1+sqrt(x))^6 ?

The integral from 0 to 1 of 3(1+sqrt(x))^6 is 2307/28

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integral from 0 to 1 of 3(1+sqrt(x))^6

\int_{0}^{1} 3 (1 + x)^{6} d x

\frac{2307}{28}

Solution steps

\int_{0}^{1} 3 (1 + x)^{6} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{1} (1 + x)^{6} d x

Expand

(1 + x)^{6} : 1 + 6 x + 15 x + 20 x^{\frac{3}{2}} + 15 x^{2} + 6 x^{\frac{5}{2}} + x^{3}

= 3 \cdot \int_{0}^{1} 1 + 6 x + 15 x + 20 x^{\frac{3}{2}} + 15 x^{2} + 6 x^{\frac{5}{2}} + x^{3} 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

= 3 (\int_{0}^{1} 1 d x + \int_{0}^{1} 6 x d x + \int_{0}^{1} 15 x d x + \int_{0}^{1} 20 x^{\frac{3}{2}} d x + \int_{0}^{1} 15 x^{2} d x + \int_{0}^{1} 6 x^{\frac{5}{2}} d x + \int_{0}^{1} x^{3} d x)

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

\int_{0}^{1} 6 x d x = 4

\int_{0}^{1} 15 x d x = \frac{15}{2}

\int_{0}^{1} 20 x^{\frac{3}{2}} d x = 8

\int_{0}^{1} 15 x^{2} d x = 5

\int_{0}^{1} 6 x^{\frac{5}{2}} d x = \frac{12}{7}

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

= 3 (1 + 4 + \frac{15}{2} + 8 + 5 + \frac{12}{7} + \frac{1}{4})

Simplify

3 (1 + 4 + \frac{15}{2} + 8 + 5 + \frac{12}{7} + \frac{1}{4}) : \frac{2307}{28}

= \frac{2307}{28}

\int r 4 - r^{2} - r^{2} d r

\frac{1}{( s )}

sin (sin (x))

\frac{d}{d x} ((x - 1)^{2} (x - 2))

f (x) = sin (x^{3})

What is the integral from 0 to 1 of 3(1+sqrt(x))^6 ?
The integral from 0 to 1 of 3(1+sqrt(x))^6 is 2307/28