Solve ∫ (from 0 to 6) of (4t-t^2)-(t^2-8t)dt

$$\int _ { 0 } ^ { 6 } ( 4 t - t ^ { 2 } ) - ( t ^ { 2 } - 8 t ) d t$$

Evaluate

$$72$$

Solution Steps

To find the opposite of t^{2}-8t, find the opposite of each term.

$$\int _{0}^{6}4t-t^{2}-t^{2}+8t\mathrm{d}t$$

Combine -t^{2} and -t^{2} to get -2t^{2}.

$$\int _{0}^{6}4t-2t^{2}+8t\mathrm{d}t$$

Combine 4t and 8t to get 12t.

$$\int _{0}^{6}12t-2t^{2}\mathrm{d}t$$

Evaluate the indefinite integral first.

$$\int 12t-2t^{2}\mathrm{d}t$$

Integrate the sum term by term.

$$\int 12t\mathrm{d}t+\int -2t^{2}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$12\int t\mathrm{d}t-2\int t^{2}\mathrm{d}t$$

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

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

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}.

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

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.

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

Simplify.

$$72$$