Solve = ∫ (from 0 to 4) of (5+8t-2t^2)dt

$$= \int _ { 0 } ^ { 4 } ( 5 + 8 t - 2 t ^ { 2 } ) d t$$

Evaluate

$$\frac{124}{3}\approx 41.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 5+8t-2t^{2}\mathrm{d}t$$

Integrate the sum term by term.

$$\int 5\mathrm{d}t+\int 8t\mathrm{d}t+\int -2t^{2}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int 5\mathrm{d}t+8\int t\mathrm{d}t-2\int t^{2}\mathrm{d}t$$

Find the integral of 5 using the table of common integrals rule \int a\mathrm{d}t=at.

$$5t+8\int t\mathrm{d}t-2\int t^{2}\mathrm{d}t$$

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

$$5t+4t^{2}-2\int t^{2}\mathrm{d}t$$

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

$$5t+4t^{2}-\frac{2t^{3}}{3}$$

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.

$$5\times 4+4\times 4^{2}-\frac{2}{3}\times 4^{3}-\left(5\times 0+4\times 0^{2}-\frac{2}{3}\times 0^{3}\right)$$

Simplify.

$$\frac{124}{3}$$