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

The integral from 0 to 4 of (x^3-x^2+1) is 140/3

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

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

\frac{140}{3}

Decimal

46.66666 \dots

Solution steps

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

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

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

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

= 64 - \frac{64}{3} + 4

Simplify

64 - \frac{64}{3} + 4 : \frac{140}{3}

= \frac{140}{3}

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What is the integral from 0 to 4 of (x^3-x^2+1) ?
The integral from 0 to 4 of (x^3-x^2+1) is 140/3