What is the integral from 0 to 1 of (-3x^2+4x+3) ?

The integral from 0 to 1 of (-3x^2+4x+3) is 4

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integral from 0 to 1 of (-3x^2+4x+3)

\int_{0}^{1} (- 3 x^{2} + 4 x + 3) d x

4

Solution steps

\int_{0}^{1} - 3 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}^{1} 3 x^{2} d x + \int_{0}^{1} 4 x d x + \int_{0}^{1} 3 d x

\int_{0}^{1} 3 x^{2} d x = 1

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

\int_{0}^{1} 3 d x = 3

= - 1 + 2 + 3

Simplify

= 4

What is the integral from 0 to 1 of (-3x^2+4x+3) ?
The integral from 0 to 1 of (-3x^2+4x+3) is 4