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

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

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

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

\frac{7}{18}

Decimal

0.38888 \dots

Solution steps

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

\frac{tan ( x ) - 20}{sec ( x )}

\int 4 (tan^{2} (x) + tan^{4} (x)) d x

y^{^{''}} - 6 y^{^{'}} + 5 y = 0

\frac{\partial}{\partial x} (5 - x^{2} + x y - 3 y^{2})

h (x) = ln (x + x^{2} - 11)

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