Solve ∫ (from 0 to 2) of (6t-3t^2)dt

$$\int_{ 0 }^{ 2 } (6t-3 { t }^{ 2 } ) d t$$

Evaluate

$$4$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 6t-3t^{2}\mathrm{d}t$$

Integrate the sum term by term.

$$\int 6t\mathrm{d}t+\int -3t^{2}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$6\int t\mathrm{d}t-3\int t^{2}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t\mathrm{d}t with \frac{t^{2}}{2}. Multiply 6 times \frac{t^{2}}{2}.

$$3t^{2}-3\int t^{2}\mathrm{d}t$$

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 -3 times \frac{t^{3}}{3}.

$$3t^{2}-t^{3}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$3\times 2^{2}-2^{3}-\left(3\times 0^{2}-0^{3}\right)$$

Simplify.

$$4$$