Solve ∫ (from - 1 to 2) of (t+1)(t-2)dt

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

Evaluate

$$-\frac{9}{2}=-4.5$$

Solution Steps

Apply the distributive property by multiplying each term of t+1 by each term of t-2.

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

Combine -2t and t to get -t.

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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 -1 times \frac{t^{2}}{2}.

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

Find the integral of -2 using the table of common integrals rule \int a\mathrm{d}t=at.

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

Simplify.

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

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.

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

Simplify.

$$-\frac{9}{2}$$