What is the integral from-1 to 2 of (x^3-7x) ?

The integral from-1 to 2 of (x^3-7x) is -27/4

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integral from-1 to 2 of (x^3-7x)

\int_{- 1}^{2} (x^{3} - 7 x) d x

- \frac{27}{4}

Decimal Notation

- 6.75

Solution steps

\int_{- 1}^{2} x^{3} - 7 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} 7 x d x

\int_{- 1}^{2} x^{3} d x = \frac{15}{4}

\int_{- 1}^{2} 7 x d x = \frac{21}{2}

= \frac{15}{4} - \frac{21}{2}

Simplify

\frac{15}{4} - \frac{21}{2} : - \frac{27}{4}

= - \frac{27}{4}

\int_{0}^{\frac{π}{2}} x \cdot sin (2 x) d x

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

\int_{0}^{\frac{π}{4}} cos^{2} (2 x) d x

\int_{- π}^{π} x^{2} cos (x) d x

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

What is the integral from-1 to 2 of (x^3-7x) ?
The integral from-1 to 2 of (x^3-7x) is -27/4