Solve ∫ (from - 1 to 1) of t(1-2t)dt

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

Evaluate

$$-\frac{4}{3}\approx -1.333333333$$

Solution Steps

Use the distributive property to multiply t by 1-2t.

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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 -2 times \frac{t^{3}}{3}.

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

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.

$$\frac{1^{2}}{2}-\frac{2}{3}\times 1^{3}-\left(\frac{\left(-1\right)^{2}}{2}-\frac{2}{3}\left(-1\right)^{3}\right)$$

Simplify.

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