Solve ∫ (t-1)dt

$$\int ( t - 1 ) d t$$

Evaluate

$$\frac{t^{2}}{2}-t+С$$

Solution Steps

Integrate the sum term by term.

$$\int t\mathrm{d}t+\int -1\mathrm{d}t$$

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}.

$$\frac{t^{2}}{2}+\int -1\mathrm{d}t$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}t=at.

$$\frac{t^{2}}{2}-t$$

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.

$$\frac{t^{2}}{2}-t+С$$

Differentiate w.r.t. t

$$t-1$$