What is the integral of $4 x^{3}$ $4x^3$ ?

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Answer 1 · 2018-03-11T19:43:38

The integral equals $x^{4} + C$ $x^4 + C$

As given by the power rule, $\int x^{n} d x = \frac{x^{n + 1}}{n + 1}$ $int x^ndx = x^(n + 1)/(n + 1)$ .

$I = 4 \frac{x^{3 + 1}}{3 + 1} = x^{4} + C$ $I = 4x^(3+ 1)/(3 + 1) = x^4 + C$

Hopefully this helps!

Answer 2 · 2018-03-11T19:45:47

$\int 4 x^{3} d x = x^{4} + C$ $\int 4x^3 dx= x^4 + C$

$\int 4 x^{3} d x$ $\int 4x^3 dx$

$= 4 (\frac{x^{3 + 1}}{3 + 1}) + C$ $=4(x^(3+1)/(3+1))+C$ ; $C \equiv$ $C\equiv$ arbitrary constant

$= 4 (\frac{x^{4}}{4}) + C$ $=4(x^(4)/(4))+C$

$= x^{4} + C$ $= x^4 + C$

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