What is the integral from 0 to 2 of (2x-6)(4x^2+5) ?

The integral from 0 to 2 of (2x-6)(4x^2+5) is -72

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integral from 0 to 2 of (2x-6)(4x^2+5)

\int_{0}^{2} (2 x - 6) (4 x^{2} + 5) d x

- 72

Solution steps

\int_{0}^{2} (2 x - 6) (4 x^{2} + 5) d x

Expand

(2 x - 6) (4 x^{2} + 5) : 8 x^{3} + 10 x - 24 x^{2} - 30

= \int_{0}^{2} 8 x^{3} + 10 x - 24 x^{2} - 30 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}^{2} 8 x^{3} d x + \int_{0}^{2} 10 x d x - \int_{0}^{2} 24 x^{2} d x - \int_{0}^{2} 30 d x

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

\int_{0}^{2} 10 x d x = 20

\int_{0}^{2} 24 x^{2} d x = 64

\int_{0}^{2} 30 d x = 60

= 32 + 20 - 64 - 60

Simplify

= - 72

\int_{2}^{5} 6 + cos (\frac{x}{10}) d x

\int_{0}^{1} 2 x + 4 d x

\int_{\frac{1}{2}}^{\frac{1}{3}} (cos (π x)) d x

\int_{0}^{1} 2 π x 1 + 4 x^{2} d x

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

What is the integral from 0 to 2 of (2x-6)(4x^2+5) ?
The integral from 0 to 2 of (2x-6)(4x^2+5) is -72