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

The integral from 1 to 2 of 4x^3-3x^2+4x+2 is 16

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integral from 1 to 2 of 4x^3-3x^2+4x+2

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

16

Solution steps

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

\int_{1}^{2} 4 x^{3} d x = 15

\int_{1}^{2} 3 x^{2} d x = 7

\int_{1}^{2} 4 x d x = 6

\int_{1}^{2} 2 d x = 2

= 15 - 7 + 6 + 2

Simplify

= 16

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\int_{0}^{\frac{1}{15}} x ln (15 x) d x

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

What is the integral from 1 to 2 of 4x^3-3x^2+4x+2 ?
The integral from 1 to 2 of 4x^3-3x^2+4x+2 is 16