# How do you find the antiderivative of f(x)=-1/2x^3+2x^2-3x-2?

Jun 2, 2018

$- \frac{1}{8} {x}^{4} + \frac{2}{3} {x}^{3} - \frac{3}{2} {x}^{2} - 2 x + C$

#### Explanation:

The antiderivative of the function is basically the integral of the function. So here we have: $\int - \frac{1}{2} {x}^{3} + 2 {x}^{2} - 3 x - 2 \setminus \mathrm{dx}$.

We use the power rule and the constant rule, which states that,

• $\int {a}^{n} \setminus \mathrm{dx} = \frac{{a}^{n + 1}}{n + 1} + C$

• $\int a \cdot f \left(x\right) \setminus \mathrm{dx} = a \int f \left(x\right) \setminus \mathrm{dx}$,

respectively.

$= - \frac{1}{8} {x}^{4} + \frac{2}{3} {x}^{3} - \frac{3}{2} {x}^{2} - 2 x + C$

Jun 2, 2018

$- \frac{1}{8} {x}^{4} + \frac{2}{3} {x}^{3} - \frac{3}{2} {x}^{2} - 2 x + c$

#### Explanation:

$\text{integrate each term using the "color(blue)"power rule}$

â€¢color(white)(x)int(ax^n)=a/(n+1)x^(n+1)ton!=-1

$\int \left(- \frac{1}{2} {x}^{3} + 2 {x}^{2} - 3 x - 2\right) \mathrm{dx}$

$= - \frac{1}{8} {x}^{4} + \frac{2}{3} {x}^{3} - \frac{3}{2} {x}^{2} - 2 x + c$

$\text{where c is the constant of integration}$