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

The integral from 0 to x of (3t^2-t-4) is x^3-(x^2)/2-4x

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

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

x^{3} - \frac{x ^{2}}{2} - 4 x

Solution steps

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

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

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

\int_{0}^{x} t d t = \frac{x ^{2}}{2}

\int_{0}^{x} 4 d t = 4 x

= x^{3} - \frac{x ^{2}}{2} - 4 x

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