Solve ∫ (from 1 to 5) of (1.5t^3-135t^2+22.5t)dt=

$$\int _ { 1 } ^ { 5 } ( 1.5 t ^ { 3 } - 135 t ^ { 2 } + 22.5 t ) d t =$$

Evaluate

$$-5076$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{3t^{3}}{2}-135t^{2}+\frac{45t}{2}\mathrm{d}t$$

Integrate the sum term by term.

$$\int \frac{3t^{3}}{2}\mathrm{d}t+\int -135t^{2}\mathrm{d}t+\int \frac{45t}{2}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\frac{3\int t^{3}\mathrm{d}t}{2}-135\int t^{2}\mathrm{d}t+\frac{45\int t\mathrm{d}t}{2}$$

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 1.5 times \frac{t^{4}}{4}.

$$\frac{3t^{4}}{8}-135\int t^{2}\mathrm{d}t+\frac{45\int t\mathrm{d}t}{2}$$

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 -135 times \frac{t^{3}}{3}.

$$\frac{3t^{4}}{8}-45t^{3}+\frac{45\int t\mathrm{d}t}{2}$$

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

$$\frac{3t^{4}}{8}-45t^{3}+\frac{45t^{2}}{4}$$

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.

$$\frac{3}{8}\times 5^{4}-45\times 5^{3}+\frac{45}{4}\times 5^{2}-\left(\frac{3}{8}\times 1^{4}-45\times 1^{3}+\frac{45}{4}\times 1^{2}\right)$$

Simplify.

$$-5076$$