What is the integral of x^3-1/2 x^2+4 ?

The integral of x^3-1/2 x^2+4 is (x^4)/4-(x^3)/6+4x+C

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integral of x^3-1/2 x^2+4

\int x^{3} - \frac{1}{2} x^{2} + 4 d x

\frac{x ^{4}}{4} - \frac{x ^{3}}{6} + 4 x + C

Solution steps

\int x^{3} - \frac{1}{2} x^{2} + 4 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 x^{3} d x - \int \frac{1}{2} x^{2} d x + \int 4 d x

\int x^{3} d x = \frac{x ^{4}}{4}

\int \frac{1}{2} x^{2} d x = \frac{x ^{3}}{6}

\int 4 d x = 4 x

= \frac{x ^{4}}{4} - \frac{x ^{3}}{6} + 4 x

Add a constant to the solution

= \frac{x ^{4}}{4} - \frac{x ^{3}}{6} + 4 x + C

f (x) = e^{x} (x^{4} + 3), (0, 3)

\frac{d}{d x} (\frac{x ^{2} - 10 x + 9}{( x - 5 ) ^{2}})

f (x) = \frac{4}{2 - x}

- \frac{36 ( x - 1 )}{( x + 3 ) ^{3}}

2 x^{2} + 5, - 2, 3

What is the integral of x^3-1/2 x^2+4 ?
The integral of x^3-1/2 x^2+4 is (x^4)/4-(x^3)/6+4x+C