What is the integral from o to 1 of 42x(1-x)^5 ?

The integral from o to 1 of 42x(1-x)^5 is 6o^7-35o^6+84o^5-105o^4+70o^3-21o^2+1

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integral from o to 1 of 42x(1-x)^5

\int_{o}^{1} 42 x (1 - x)^{5} d x

6 o^{7} - 35 o^{6} + 84 o^{5} - 105 o^{4} + 70 o^{3} - 21 o^{2} + 1

Solution steps

\int_{o}^{1} 42 x (1 - x)^{5} d x

Take the constant out:

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

= 42 \cdot \int_{o}^{1} x (1 - x)^{5} d x

Apply Integration By Parts

: [\frac{1}{2} x^{2} (1 - x)^{5} - \int - \frac{5}{2} x^{2} (1 - x)^{4} d x]_{o}^{1}

= 42 [\frac{1}{2} x^{2} (1 - x)^{5} - \int - \frac{5}{2} x^{2} (1 - x)^{4} d x]_{o}^{1}

\int - \frac{5}{2} x^{2} (1 - x)^{4} d x = - \frac{1}{42} (35 x^{3} - 105 x^{4} + 126 x^{5} - 70 x^{6} + 15 x^{7})

= 42 [\frac{1}{2} x^{2} (1 - x)^{5} - (- \frac{1}{42} (35 x^{3} - 105 x^{4} + 126 x^{5} - 70 x^{6} + 15 x^{7}))]_{o}^{1}

Simplify

= 42 [\frac{1}{2} x^{2} (1 - x)^{5} + \frac{1}{42} (35 x^{3} - 105 x^{4} + 126 x^{5} - 70 x^{6} + 15 x^{7})]_{o}^{1}

Compute the boundaries:

\frac{1}{42} - \frac{1}{2} o^{2} (1 - o)^{5} - \frac{5}{6} o^{3} + \frac{5}{2} o^{4} - 3 o^{5} + \frac{5}{3} o^{6} - \frac{5}{14} o^{7}

= 42 (\frac{1}{42} - \frac{1}{2} o^{2} (1 - o)^{5} - \frac{5}{6} o^{3} + \frac{5}{2} o^{4} - 3 o^{5} + \frac{5}{3} o^{6} - \frac{5}{14} o^{7})

Simplify

= 6 o^{7} - 35 o^{6} + 84 o^{5} - 105 o^{4} + 70 o^{3} - 21 o^{2} + 1

\frac{d}{d x} (a^{2} + x^{2})

x \to 1 lim (\frac{1 + x}{1 - x ^{2}})

\int_{0}^{- \frac{π}{6}} x cos (3 x) d x

y = 8 x + 7

t \to 1 lim (t + 8)

What is the integral from o to 1 of 42x(1-x)^5 ?
The integral from o to 1 of 42x(1-x)^5 is 6o^7-35o^6+84o^5-105o^4+70o^3-21o^2+1