What is the integral from 0 to 1 of x^2(x+1)^5 ?

The integral from 0 to 1 of x^2(x+1)^5 is 341/56

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integral from 0 to 1 of x^2(x+1)^5

\int_{0}^{1} x^{2} (x + 1)^{5} d x

\frac{341}{56}

Decimal

6.08928 \dots

Solution steps

\int_{0}^{1} x^{2} (x + 1)^{5} d x

Expand

x^{2} (x + 1)^{5} : x^{7} + 5 x^{6} + 10 x^{5} + 10 x^{4} + 5 x^{3} + x^{2}

= \int_{0}^{1} x^{7} + 5 x^{6} + 10 x^{5} + 10 x^{4} + 5 x^{3} + 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

= \int_{0}^{1} x^{7} d x + \int_{0}^{1} 5 x^{6} d x + \int_{0}^{1} 10 x^{5} d x + \int_{0}^{1} 10 x^{4} d x + \int_{0}^{1} 5 x^{3} d x + \int_{0}^{1} x^{2} d x

\int_{0}^{1} x^{7} d x = \frac{1}{8}

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

\int_{0}^{1} 10 x^{5} d x = \frac{5}{3}

\int_{0}^{1} 10 x^{4} d x = 2

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

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

= \frac{1}{8} + \frac{5}{7} + \frac{5}{3} + 2 + \frac{5}{4} + \frac{1}{3}

Simplify

\frac{1}{8} + \frac{5}{7} + \frac{5}{3} + 2 + \frac{5}{4} + \frac{1}{3} : \frac{341}{56}

= \frac{341}{56}

\int_{- 32}^{2} \frac{1}{∣ 2 x ∣} d x

y^{^{'}} = \frac{( y - x y )}{x ^{2}}

ln (3 x)

\int 4 x ln (y) d x

- 3 x^{3} - 2 x^{2} - 1, at x = - 1

What is the integral from 0 to 1 of x^2(x+1)^5 ?
The integral from 0 to 1 of x^2(x+1)^5 is 341/56