What is the integral from 0 to 4 of x^3-3x^2+6 ?

The integral from 0 to 4 of x^3-3x^2+6 is 24

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integral from 0 to 4 of x^3-3x^2+6

\int_{0}^{4} x^{3} - 3 x^{2} + 6 d x

24

Solution steps

\int_{0}^{4} x^{3} - 3 x^{2} + 6 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}^{4} x^{3} d x - \int_{0}^{4} 3 x^{2} d x + \int_{0}^{4} 6 d x

\int_{0}^{4} x^{3} d x = 64

\int_{0}^{4} 3 x^{2} d x = 64

\int_{0}^{4} 6 d x = 24

= 64 - 64 + 24

Simplify

= 24

\int_{0}^{x} x^{2} sin (t^{2}) d t

\int_{0}^{4} x^{3} x^{2} + 9 d x

\int_{- i}^{i} \frac{1}{z} d z

\int_{0}^{π} 2 + x^{2} d x

\int_{0}^{2 π} \frac{1}{3} d θ

What is the integral from 0 to 4 of x^3-3x^2+6 ?
The integral from 0 to 4 of x^3-3x^2+6 is 24