Solve ∫ (from 0 to 3) of (4/3t^3-3/2t^2+2/3t)dt

$$\int _ { 0 } ^ { 3 } ( \frac { 4 } { 3 } t ^ { 3 } - \frac { 3 } { 2 } t ^ { 2 } + \frac { 2 } { 3 } t ) d t$$

Evaluate

$$\frac{33}{2}=16.5$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{3}\mathrm{d}t with \frac{t^{4}}{4}. Multiply \frac{4}{3} times \frac{t^{4}}{4}.

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

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

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

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

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

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.

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

Simplify.

$$\frac{33}{2}$$