Solve ∫ (from 1 to 5) of 27t^3+162t^2+324t+216dt

$$\int _ { 1 } ^ { 5 } 27 t ^ { 3 } + 162 t ^ { 2 } + 324 t + 216 d t$$

Evaluate

$$15660$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 27t^{3}+162t^{2}+324t+216\mathrm{d}t$$

Integrate the sum term by term.

$$\int 27t^{3}\mathrm{d}t+\int 162t^{2}\mathrm{d}t+\int 324t\mathrm{d}t+\int 216\mathrm{d}t$$

Factor out the constant in each of the terms.

$$27\int t^{3}\mathrm{d}t+162\int t^{2}\mathrm{d}t+324\int t\mathrm{d}t+\int 216\mathrm{d}t$$

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

$$\frac{27t^{4}}{4}+162\int t^{2}\mathrm{d}t+324\int t\mathrm{d}t+\int 216\mathrm{d}t$$

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

$$\frac{27t^{4}}{4}+54t^{3}+324\int t\mathrm{d}t+\int 216\mathrm{d}t$$

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

$$\frac{27t^{4}}{4}+54t^{3}+162t^{2}+\int 216\mathrm{d}t$$

Find the integral of 216 using the table of common integrals rule \int a\mathrm{d}t=at.

$$\frac{27t^{4}}{4}+54t^{3}+162t^{2}+216t$$

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.

$$\frac{27}{4}\times 5^{4}+54\times 5^{3}+162\times 5^{2}+216\times 5-\left(\frac{27}{4}\times 1^{4}+54\times 1^{3}+162\times 1^{2}+216\times 1\right)$$

Simplify.

$$15660$$