What is the integral from-1 to 1 of (7x^3-2x^2+5x-4) ?

The integral from-1 to 1 of (7x^3-2x^2+5x-4) is -28/3

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integral from-1 to 1 of (7x^3-2x^2+5x-4)

\int_{- 1}^{1} (7 x^{3} - 2 x^{2} + 5 x - 4) d x

- \frac{28}{3}

Decimal Notation

- 9.33333 \dots

Solution steps

\int_{- 1}^{1} 7 x^{3} - 2 x^{2} + 5 x - 4 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}^{1} 7 x^{3} d x - \int_{- 1}^{1} 2 x^{2} d x + \int_{- 1}^{1} 5 x d x - \int_{- 1}^{1} 4 d x

\int_{- 1}^{1} 7 x^{3} d x = 0

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

\int_{- 1}^{1} 5 x d x = 0

\int_{- 1}^{1} 4 d x = 8

= 0 - \frac{4}{3} + 0 - 8

Simplify

0 - \frac{4}{3} + 0 - 8 : - \frac{28}{3}

= - \frac{28}{3}

\frac{d y}{d x} = \frac{2 x + 1}{2 y}

f (x) = \frac{1}{x + 3}

(ln (x))^{^{'}}

\frac{d y}{d x} = (- 3 y)

\int 169 - x^{2} d x

What is the integral from-1 to 1 of (7x^3-2x^2+5x-4) ?
The integral from-1 to 1 of (7x^3-2x^2+5x-4) is -28/3