What is [(x^4)/4-x^3-(x^2)/2+3x]_{0}^1 ?

The solution to [(x^4)/4-x^3-(x^2)/2+3x]_{0}^1 is (x^4)/4-x^3-(x^2)/2+3x

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simplify [(x^4)/4-x^3-(x^2)/2+3x]_{0}^1

simplify

[\frac{x ^{4}}{4} - x^{3} - \frac{x ^{2}}{2} + 3 x]_{0}^{1}

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

Solution steps

(\frac{x ^{4}}{4} - x^{3} - \frac{x ^{2}}{2} + 3 x)^{1}

Apply exponent rule:

a^{1} = a

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

(d^{- 5})^{- 2}

lo g_{10} (x) (x - 2)

x^{2} y + x y^{2} - x - y

x^{3} + x^{2} - 2 x + 3 + 3 x - 2

4 a^{8} \cdot b^{c} - 6 a^{5} \cdot b^{3}

What is [(x^4)/4-x^3-(x^2)/2+3x]_{0}^1 ?
The solution to [(x^4)/4-x^3-(x^2)/2+3x]_{0}^1 is (x^4)/4-x^3-(x^2)/2+3x