What is the integral of-0.6x^2+80 ?

The integral of-0.6x^2+80 is -0.2x^3+80x+C

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integral of-0.6x^2+80

\int - 0.6 x^{2} + 80 d x

- 0.2 x^{3} + 80 x + C

Solution steps

\int - 0.6 x^{2} + 80 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 0.6 x^{2} d x + \int 80 d x

\int 0.6 x^{2} d x = 0.2 x^{3}

\int 80 d x = 80 x

= - 0.2 x^{3} + 80 x

Add a constant to the solution

= - 0.2 x^{3} + 80 x + C

\int θ sec^{2} (θ) d θ

1 + 2 x

\int \frac{1}{t ^{2} - 1} d t

\frac{d}{d x} (x^{2} + \frac{y ^{2}}{9}) = 1

t^{\frac{3}{2}}