What is the integral from 3 to 0 of (-x^3-2x^2+x-1) ?

The integral from 3 to 0 of (-x^3-2x^2+x-1) is 147/4

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integral from 3 to 0 of (-x^3-2x^2+x-1)

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

\frac{147}{4}

Decimal

36.75

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{3} - x^{3} - 2 x^{2} + x - 1 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_{0}^{3} x^{3} d x - \int_{0}^{3} 2 x^{2} d x + \int_{0}^{3} x d x - \int_{0}^{3} 1 d x)

\int_{0}^{3} x^{3} d x = \frac{81}{4}

\int_{0}^{3} 2 x^{2} d x = 18

\int_{0}^{3} x d x = \frac{9}{2}

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

= - (- \frac{81}{4} - 18 + \frac{9}{2} - 3)

Simplify

- (- \frac{81}{4} - 18 + \frac{9}{2} - 3) : \frac{147}{4}

= \frac{147}{4}

y = \frac{x}{x + 3}, (1, 0.25)

\frac{d}{d x} (\frac{1}{( 3 x ^{2} + x - 3 ) ^{9}})

f (x) = x^{x^{2}}

x \to 9 lim (25 - x)

\frac{\partial}{\partial x} (2 x y e^{x^{2}})

What is the integral from 3 to 0 of (-x^3-2x^2+x-1) ?
The integral from 3 to 0 of (-x^3-2x^2+x-1) is 147/4