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

The integral from 2 to 4 of (2x-3)(3x+1) is 64

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

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

64

Solution steps

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

Expand

(2 x - 3) (3 x + 1) : 6 x^{2} - 7 x - 3

= \int_{2}^{4} 6 x^{2} - 7 x - 3 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}^{4} 6 x^{2} d x - \int_{2}^{4} 7 x d x - \int_{2}^{4} 3 d x

\int_{2}^{4} 6 x^{2} d x = 112

\int_{2}^{4} 7 x d x = 42

\int_{2}^{4} 3 d x = 6

= 112 - 42 - 6

Simplify

= 64

\frac{1}{y ^{2}} \cdot (y \cdot x + x^{2})

\int 2 r^{2} d r

x cos (1 - e^{x})

\int (\frac{7}{x ^{5}} + 4 x + 3) d x

ln (x x)

What is the integral from 2 to 4 of (2x-3)(3x+1) ?
The integral from 2 to 4 of (2x-3)(3x+1) is 64