Solve = ∫ (from 0 to 1) of (12+6t)dt

$$= \int _ { 0 } ^ { 1 } ( 12 + 6 t ) d t$$

Evaluate

$$15$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 12+6t\mathrm{d}t$$

Integrate the sum term by term.

$$\int 12\mathrm{d}t+\int 6t\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int 12\mathrm{d}t+6\int t\mathrm{d}t$$

Find the integral of 12 using the table of common integrals rule \int a\mathrm{d}t=at.

$$12t+6\int t\mathrm{d}t$$

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}.

$$12t+3t^{2}$$

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.

$$12\times 1+3\times 1^{2}-\left(12\times 0+3\times 0^{2}\right)$$

Simplify.

$$15$$