Solve ∫ 20t-20dt

$$\int 20 t - 20 d t$$

Evaluate

$$10t^{2}-20t+С$$

Solution Steps

Integrate the sum term by term.

$$\int 20t\mathrm{d}t+\int -20\mathrm{d}t$$

Factor out the constant in each of the terms.

$$20\int t\mathrm{d}t+\int -20\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t\mathrm{d}t with \frac{t^{2}}{2}. Multiply 20 times \frac{t^{2}}{2}.

$$10t^{2}+\int -20\mathrm{d}t$$

Find the integral of -20 using the table of common integrals rule \int a\mathrm{d}t=at.

$$10t^{2}-20t$$

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

$$10t^{2}-20t+С$$

Differentiate w.r.t. t

$$20\left(t-1\right)$$