What is the integral from-1 to 1 of t(1-t)^2 ?

The integral from-1 to 1 of t(1-t)^2 is -4/3

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integral from-1 to 1 of t(1-t)^2

\int_{- 1}^{1} t (1 - t)^{2} d t

- \frac{4}{3}

Decimal Notation

- 1.33333 \dots

Solution steps

\int_{- 1}^{1} t (1 - t)^{2} d t

Expand

t (1 - t)^{2} : t - 2 t^{2} + t^{3}

= \int_{- 1}^{1} t - 2 t^{2} + t^{3} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{1} t d t - \int_{- 1}^{1} 2 t^{2} d t + \int_{- 1}^{1} t^{3} d t

\int_{- 1}^{1} t d t = 0

\int_{- 1}^{1} 2 t^{2} d t = \frac{4}{3}

\int_{- 1}^{1} t^{3} d t = 0

= 0 - \frac{4}{3} + 0

Simplify

= - \frac{4}{3}

x \to 0 lim (x^{3} ln (x^{2} + 1))

n = 1 \sum \infty \frac{( - 3 ) ^{n - 1}}{8 ^{n}}

\frac{\partial}{\partial x} (4 (x - 1)^{2})

n = 1 \sum \infty n sin (\frac{1}{n})

\frac{\partial}{\partial x} (2 + 3 x + 4 y)

What is the integral from-1 to 1 of t(1-t)^2 ?
The integral from-1 to 1 of t(1-t)^2 is -4/3