What is the integral from 0 to 1 of 5(1+sqrt(x))^4 ?

The integral from 0 to 1 of 5(1+sqrt(x))^4 is 43

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integral from 0 to 1 of 5(1+sqrt(x))^4

\int_{0}^{1} 5 (1 + x)^{4} d x

43

Solution steps

\int_{0}^{1} 5 (1 + x)^{4} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} (1 + x)^{4} d x

Expand

(1 + x)^{4} : 1 + 4 x + 6 x + 4 x^{\frac{3}{2}} + x^{2}

= 5 \cdot \int_{0}^{1} 1 + 4 x + 6 x + 4 x^{\frac{3}{2}} + x^{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

= 5 (\int_{0}^{1} 1 d x + \int_{0}^{1} 4 x d x + \int_{0}^{1} 6 x d x + \int_{0}^{1} 4 x^{\frac{3}{2}} d x + \int_{0}^{1} x^{2} d x)

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

\int_{0}^{1} 4 x d x = \frac{8}{3}

\int_{0}^{1} 6 x d x = 3

\int_{0}^{1} 4 x^{\frac{3}{2}} d x = \frac{8}{5}

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

= 5 (1 + \frac{8}{3} + 3 + \frac{8}{5} + \frac{1}{3})

Simplify

5 (1 + \frac{8}{3} + 3 + \frac{8}{5} + \frac{1}{3}) : 43

= 43

\frac{\partial}{\partial x} (tan (x^{2} - y^{2}))

\frac{\partial}{\partial x} (20 x + 80 y - x^{2} - 2 y^{2})

\frac{d}{d x} (sin^{7} (2 x))

\frac{d}{d x} (- 10 e^{- 2 x})

y = \frac{6 x}{7 - tan ( x )}

What is the integral from 0 to 1 of 5(1+sqrt(x))^4 ?
The integral from 0 to 1 of 5(1+sqrt(x))^4 is 43