Solve ∫ (from 3 to 4) of (t^2+t)dt

$$\int _ { 3 } ^ { 4 } ( t ^ { 2 } + t ) d t$$

Evaluate

$$\frac{95}{6}\approx 15.833333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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}.

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

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}.

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

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.

$$\frac{4^{3}}{3}+\frac{4^{2}}{2}-\left(\frac{3^{3}}{3}+\frac{3^{2}}{2}\right)$$

Simplify.

$$\frac{95}{6}$$