Solve ∫ (from 1 to 5) of (15t^3-135t^2+225t)dt=

$$\int _ { 1 } ^ { 5 } ( 15 t ^ { 3 } - 135 t ^ { 2 } + 225 t ) d t =$$

Evaluate

$$-540$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 15t^{3}-135t^{2}+225t\mathrm{d}t$$

Integrate the sum term by term.

$$\int 15t^{3}\mathrm{d}t+\int -135t^{2}\mathrm{d}t+\int 225t\mathrm{d}t$$

Factor out the constant in each of the terms.

$$15\int t^{3}\mathrm{d}t-135\int t^{2}\mathrm{d}t+225\int t\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{3}\mathrm{d}t with \frac{t^{4}}{4}. Multiply 15 times \frac{t^{4}}{4}.

$$\frac{15t^{4}}{4}-135\int t^{2}\mathrm{d}t+225\int t\mathrm{d}t$$

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

$$\frac{15t^{4}}{4}-45t^{3}+225\int t\mathrm{d}t$$

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}. Multiply 225 times \frac{t^{2}}{2}.

$$\frac{15t^{4}}{4}-45t^{3}+\frac{225t^{2}}{2}$$

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.

$$\frac{15}{4}\times 5^{4}-45\times 5^{3}+\frac{225}{2}\times 5^{2}-\left(\frac{15}{4}\times 1^{4}-45\times 1^{3}+\frac{225}{2}\times 1^{2}\right)$$

Simplify.

$$-540$$