Solve ∫ [18x^2-20x^3-1/2x^4) wrt x

$$\int [ 18 x ^ { 2 } - 20 x ^ { 3 } - \frac { 1 } { 2 } x ^ { 4 } ) d x$$

Evaluate

$$-\frac{x^{5}}{10}-5x^{4}+6x^{3}+С$$

Solution Steps

Integrate the sum term by term.

$$\int 18x^{2}\mathrm{d}x+\int -20x^{3}\mathrm{d}x+\int -\frac{x^{4}}{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$18\int x^{2}\mathrm{d}x-20\int x^{3}\mathrm{d}x-\frac{\int x^{4}\mathrm{d}x}{2}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 18 times \frac{x^{3}}{3}.

$$6x^{3}-20\int x^{3}\mathrm{d}x-\frac{\int x^{4}\mathrm{d}x}{2}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -20 times \frac{x^{4}}{4}.

$$6x^{3}-5x^{4}-\frac{\int x^{4}\mathrm{d}x}{2}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply -\frac{1}{2} times \frac{x^{5}}{5}.

$$6x^{3}-5x^{4}-\frac{x^{5}}{10}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$6x^{3}-5x^{4}-\frac{x^{5}}{10}+С$$

Differentiate w.r.t. x

$$\frac{x^{2}\left(36-40x-x^{2}\right)}{2}$$