Solve ∫ (-32t+69)}dt

$$\int{ (-32t+69) }d t$$

Evaluate

$$С+69t-16t^{2}$$

Solution Steps

Integrate the sum term by term.

$$\int -32t\mathrm{d}t+\int 69\mathrm{d}t$$

Factor out the constant in each of the terms.

$$-32\int t\mathrm{d}t+\int 69\mathrm{d}t$$

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 -32 times \frac{t^{2}}{2}.

$$-16t^{2}+\int 69\mathrm{d}t$$

Find the integral of 69 using the table of common integrals rule \int a\mathrm{d}t=at.

$$-16t^{2}+69t$$

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.

$$-16t^{2}+69t+С$$

Differentiate w.r.t. t

$$69-32t$$