What is the integral from-x^3 to x of (t^2-6t) ?

The integral from-x^3 to x of (t^2-6t) is (x^3+x^9)/3-3(-x^6+x^2)

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integral from-x^3 to x of (t^2-6t)

\int_{- x^{3}}^{x} (t^{2} - 6 t) d t

\frac{x ^{3} + x ^{9}}{3} - 3 (- x^{6} + x^{2})

Solution steps

\int_{- x^{3}}^{x} t^{2} - 6 t 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_{- x^{3}}^{x} t^{2} d t - \int_{- x^{3}}^{x} 6 t d t

\int_{- x^{3}}^{x} t^{2} d t = \frac{x ^{3} + x ^{9}}{3}

\int_{- x^{3}}^{x} 6 t d t = 3 (- x^{6} + x^{2})

= \frac{x ^{3} + x ^{9}}{3} - 3 (- x^{6} + x^{2})

\int_{- 2}^{- 2} x d x

\int_{- 1}^{1} x^{\frac{1}{3}} - cos (x) d x

\int_{- 1}^{1} k \cdot x^{2} d x

\int_{0}^{4} 5 - x^{2} d x

\int_{0}^{1} y^{3} (1 - y)^{2} d y

What is the integral from-x^3 to x of (t^2-6t) ?
The integral from-x^3 to x of (t^2-6t) is (x^3+x^9)/3-3(-x^6+x^2)