What is the integral from 0 to 2 of 12x^2+4x-3 ?

The integral from 0 to 2 of 12x^2+4x-3 is 34

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integral from 0 to 2 of 12x^2+4x-3

\int_{0}^{2} 12 x^{2} + 4 x - 3 d x

34

Solution steps

\int_{0}^{2} 12 x^{2} + 4 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_{0}^{2} 12 x^{2} d x + \int_{0}^{2} 4 x d x - \int_{0}^{2} 3 d x

\int_{0}^{2} 12 x^{2} d x = 32

\int_{0}^{2} 4 x d x = 8

\int_{0}^{2} 3 d x = 6

= 32 + 8 - 6

Simplify

= 34

\int_{0}^{4} (x - 3) (x - 5) d x

\int_{1}^{4} \frac{1}{x + 1} d x

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

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

\int_{0}^{1} (4 x - 3 x^{3}) d x

What is the integral from 0 to 2 of 12x^2+4x-3 ?
The integral from 0 to 2 of 12x^2+4x-3 is 34