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

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

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integral of 1/2+3/4 x^2-4/5 x^3

\int \frac{1}{2} + \frac{3}{4} x^{2} - \frac{4}{5} x^{3} d x

\frac{1}{2} x + \frac{x ^{3}}{4} - \frac{x ^{4}}{5} + C

Solution steps

\int \frac{1}{2} + \frac{3}{4} x^{2} - \frac{4}{5} x^{3} 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 \frac{1}{2} d x + \int \frac{3}{4} x^{2} d x - \int \frac{4}{5} x^{3} d x

\int \frac{1}{2} d x = \frac{1}{2} x

\int \frac{3}{4} x^{2} d x = \frac{x ^{3}}{4}

\int \frac{4}{5} x^{3} d x = \frac{x ^{4}}{5}

= \frac{1}{2} x + \frac{x ^{3}}{4} - \frac{x ^{4}}{5}

Add a constant to the solution

= \frac{1}{2} x + \frac{x ^{3}}{4} - \frac{x ^{4}}{5} + C

3 e^{(- 2 t)}

y = \frac{4}{x}, y = 8 x, y = \frac{1}{8} x

\frac{π}{9} cos (θ)

\int_{0}^{1} x^{2} 1 - x^{2} d x

f (x) = x sin (2 x)

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