Solve ∫ (from 0 to 2) of (6t+3)dt

$$\int _ { 0 } ^ { 2 } ( 6 t + 3 ) d t$$

Evaluate

$$18$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 6t+3\mathrm{d}t$$

Integrate the sum term by term.

$$\int 6t\mathrm{d}t+\int 3\mathrm{d}t$$

Factor out the constant in each of the terms.

$$6\int t\mathrm{d}t+\int 3\mathrm{d}t$$

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}.

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

Find the integral of 3 using the table of common integrals rule \int a\mathrm{d}t=at.

$$3t^{2}+3t$$

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.

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

Simplify.

$$18$$