What is the integral from 0 to 1 of x^2((2(x+1))/3) ?

The integral from 0 to 1 of x^2((2(x+1))/3) is 7/18

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integral from 0 to 1 of x^2((2(x+1))/3)

\int_{0}^{1} x^{2} (\frac{2 ( x + 1 )}{3}) d x

\frac{7}{18}

Decimal

0.38888 \dots

Solution steps

\int_{0}^{1} x^{2} \frac{2 ( x + 1 )}{3} d x

Take the constant out:

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

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

Expand

x^{2} (x + 1) : x^{3} + x^{2}

= \frac{2}{3} \cdot \int_{0}^{1} 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

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

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

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

= \frac{2}{3} (\frac{1}{4} + \frac{1}{3})

Simplify

\frac{2}{3} (\frac{1}{4} + \frac{1}{3}) : \frac{7}{18}

= \frac{7}{18}

x \to π - lim (ln (x))

\int_{0}^{4} 2 π x (4 x - x^{2}) d x

y^{4} - 4 y^{2} = x^{4} - 9 x^{2}, (3, - 2)

\int x^{2} (1 - x)^{6} d x

\frac{d}{d x} (- \frac{1}{x ^{7.7}})

What is the integral from 0 to 1 of x^2((2(x+1))/3) ?
The integral from 0 to 1 of x^2((2(x+1))/3) is 7/18