Solve ∫ 400x-1/2x^2-20000-100x

$$\int 400 x - \frac { 1 } { 2 } x ^ { 2 } - 20000 - 100 x$$

Evaluate

$$-\frac{x^{3}}{6}+150x^{2}-20000x+С$$

Solution Steps

Integrate the sum term by term.

$$\int 400x\mathrm{d}x+\int -\frac{x^{2}}{2}\mathrm{d}x+\int -20000\mathrm{d}x+\int -100x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$400\int x\mathrm{d}x-\frac{\int x^{2}\mathrm{d}x}{2}+\int -20000\mathrm{d}x-100\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 400 times \frac{x^{2}}{2}.

$$200x^{2}-\frac{\int x^{2}\mathrm{d}x}{2}+\int -20000\mathrm{d}x-100\int x\mathrm{d}x$$

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 -\frac{1}{2} times \frac{x^{3}}{3}.

$$200x^{2}-\frac{x^{3}}{6}+\int -20000\mathrm{d}x-100\int x\mathrm{d}x$$

Find the integral of -20000 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$200x^{2}-\frac{x^{3}}{6}-20000x-100\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -100 times \frac{x^{2}}{2}.

$$200x^{2}-\frac{x^{3}}{6}-20000x-50x^{2}$$

Simplify.

$$150x^{2}-\frac{x^{3}}{6}-20000x$$

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.

$$150x^{2}-\frac{x^{3}}{6}-20000x+С$$

Differentiate w.r.t. x

$$-\frac{x^{2}}{2}+300x-20000$$