Solve ∫ (from 0 to 3) of (t^2+3t-6)dt

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

Evaluate

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

Solution Steps

Evaluate the indefinite integral first.

$$\int t^{2}+3t-6\mathrm{d}t$$

Integrate the sum term by term.

$$\int t^{2}\mathrm{d}t+\int 3t\mathrm{d}t+\int -6\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int t^{2}\mathrm{d}t+3\int t\mathrm{d}t+\int -6\mathrm{d}t$$

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}.

$$\frac{t^{3}}{3}+3\int t\mathrm{d}t+\int -6\mathrm{d}t$$

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

$$\frac{t^{3}}{3}+\frac{3t^{2}}{2}+\int -6\mathrm{d}t$$

Find the integral of -6 using the table of common integrals rule \int a\mathrm{d}t=at.

$$\frac{t^{3}}{3}+\frac{3t^{2}}{2}-6t$$

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.

$$\frac{3^{3}}{3}+\frac{3}{2}\times 3^{2}-6\times 3-\left(\frac{0^{3}}{3}+\frac{3}{2}\times 0^{2}-6\times 0\right)$$

Simplify.

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