Solve ∫ (from 0 to 3) of (3t^2-8t-3)dt

$$\int _ { 0 } ^ { 3 } ( 3 t ^ { 2 } - 8 t - 3 ) d t$$

Evaluate

$$-18$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 3t^{2}-8t-3\mathrm{d}t$$

Integrate the sum term by term.

$$\int 3t^{2}\mathrm{d}t+\int -8t\mathrm{d}t+\int -3\mathrm{d}t$$

Factor out the constant in each of the terms.

$$3\int t^{2}\mathrm{d}t-8\int t\mathrm{d}t+\int -3\mathrm{d}t$$

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

$$t^{3}-8\int t\mathrm{d}t+\int -3\mathrm{d}t$$

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 -8 times \frac{t^{2}}{2}.

$$t^{3}-4t^{2}+\int -3\mathrm{d}t$$

Find the integral of -3 using the table of common integrals rule \int a\mathrm{d}t=at.

$$t^{3}-4t^{2}-3t$$

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.

$$3^{3}-4\times 3^{2}-3\times 3-\left(0^{3}-4\times 0^{2}-3\times 0\right)$$

Simplify.

$$-18$$