What is the integral from-2 to 0 of (8x^3-7x^2) ?

The integral from-2 to 0 of (8x^3-7x^2) is -152/3

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integral from-2 to 0 of (8x^3-7x^2)

\int_{- 2}^{0} (8 x^{3} - 7 x^{2}) d x

- \frac{152}{3}

Decimal

- 50.66666 \dots

Solution steps

\int_{- 2}^{0} 8 x^{3} - 7 x^{2} 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_{- 2}^{0} 8 x^{3} d x - \int_{- 2}^{0} 7 x^{2} d x

\int_{- 2}^{0} 8 x^{3} d x = - 32

\int_{- 2}^{0} 7 x^{2} d x = \frac{56}{3}

= - 32 - \frac{56}{3}

Simplify

- 32 - \frac{56}{3} : - \frac{152}{3}

= - \frac{152}{3}

\int_{- 2}^{2} 3 - ∣ x ∣ d x

\int_{0}^{9} x + 1 d x

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

\int_{1}^{9} (\frac{4}{x} - e^{- x}) d x

\int_{1}^{e} \frac{( x ^{5} + 1 )}{x} d x

What is the integral from-2 to 0 of (8x^3-7x^2) ?
The integral from-2 to 0 of (8x^3-7x^2) is -152/3