What is the integral of (x^2+5x(-1)/2) ?

The integral of (x^2+5x(-1)/2) is (x^3)/3-(5x^2)/4+C

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integral of (x^2+5x(-1)/2)

\int (x^{2} + 5 x \frac{- 1}{2}) d x

\frac{x ^{3}}{3} - \frac{5 x ^{2}}{4} + C

Solution steps

\int x^{2} + 5 x \frac{- 1}{2} 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^{2} d x + \int 5 x \frac{- 1}{2} d x

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

\int 5 x \frac{- 1}{2} d x = - \frac{5 x ^{2}}{4}

= \frac{x ^{3}}{3} - \frac{5 x ^{2}}{4}

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{5 x ^{2}}{4} + C

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

y^{^{'}} + 2 y = 3 x, y (0) = 3

\int 1 + sin (x) d x

x \to π lim (- 1)

\frac{\partial}{\partial x} (\frac{a ^{3} b ^{4}}{c ^{2} x ^{4}})

What is the integral of (x^2+5x(-1)/2) ?
The integral of (x^2+5x(-1)/2) is (x^3)/3-(5x^2)/4+C