What is the integral from 1 to sqrt(2 of)(x^3-6x) ?

The integral from 1 to sqrt(2 of)(x^3-6x) is -9/4

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integral from 1 to sqrt(2 of)(x^3-6x)

\int_{1}^{2} (x^{3} - 6 x) d x

- \frac{9}{4}

Decimal Notation

- 2.25

Solution steps

\int_{1}^{2} x^{3} - 6 x 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_{1}^{2} x^{3} d x - \int_{1}^{2} 6 x d x

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

\int_{1}^{2} 6 x d x = 3

= \frac{3}{4} - 3

Simplify

\frac{3}{4} - 3 : - \frac{9}{4}

= - \frac{9}{4}

\int_{0}^{1} 4 x - 3 x^{2} - x^{3} d x

\int_{0}^{2} 2 x + x d x

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

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

\int_{- 3}^{0} x^{2} + 6 x + 9 d x

What is the integral from 1 to sqrt(2 of)(x^3-6x) ?
The integral from 1 to sqrt(2 of)(x^3-6x) is -9/4