Solve ∫ (from 3 to 7) of (t^2-t-20)dt

$$\int _ { 3 } ^ { 7 } ( t ^ { 2 } - t - 20 ) d t$$

Evaluate

$$\frac{16}{3}\approx 5.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int t^{2}-t-20\mathrm{d}t$$

Integrate the sum term by term.

$$\int t^{2}\mathrm{d}t+\int -t\mathrm{d}t+\int -20\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int t^{2}\mathrm{d}t-\int t\mathrm{d}t+\int -20\mathrm{d}t$$

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}.

$$\frac{t^{3}}{3}-\int t\mathrm{d}t+\int -20\mathrm{d}t$$

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}.

$$\frac{t^{3}}{3}-\frac{t^{2}}{2}+\int -20\mathrm{d}t$$

Find the integral of -20 using the table of common integrals rule \int a\mathrm{d}t=at.

$$\frac{t^{3}}{3}-\frac{t^{2}}{2}-20t$$

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.

$$\frac{7^{3}}{3}-\frac{7^{2}}{2}-20\times 7-\left(\frac{3^{3}}{3}-\frac{3^{2}}{2}-20\times 3\right)$$

Simplify.

$$\frac{16}{3}$$