Solve ∫ 4-t^2dt

$$\int 4 - t ^ { 2 } d t$$

Evaluate

$$-\frac{t^{3}}{3}+4t+С$$

Solution Steps

Integrate the sum term by term.

$$\int 4\mathrm{d}t+\int -t^{2}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int 4\mathrm{d}t-\int t^{2}\mathrm{d}t$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}t=at.

$$4t-\int t^{2}\mathrm{d}t$$

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

$$4t-\frac{t^{3}}{3}$$

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.

$$4t-\frac{t^{3}}{3}+С$$

Differentiate w.r.t. t

$$4-t^{2}$$