Solve ∫ (from - 1 to 1) of (2t-1)dt

$$\int _ { - 1 } ^ { 1 } ( 2 t - 1 ) d t$$

Evaluate

$$-2$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2t-1\mathrm{d}t$$

Integrate the sum term by term.

$$\int 2t\mathrm{d}t+\int -1\mathrm{d}t$$

Factor out the constant in each of the terms.

$$2\int t\mathrm{d}t+\int -1\mathrm{d}t$$

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

$$t^{2}+\int -1\mathrm{d}t$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}t=at.

$$t^{2}-t$$

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.

$$1^{2}-1-\left(\left(-1\right)^{2}-\left(-1\right)\right)$$

Simplify.

$$-2$$