Solve ∫ (from 1 to 2) of (3t+5)^3dt

$$\int _ { 1 } ^ { 2 } ( 3 t + 5 ) ^ { 3 } d t$$

Evaluate

$$\frac{3515}{4}=878.75$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(3t+5\right)^{3}.

$$\int _{1}^{2}27t^{3}+135t^{2}+225t+125\mathrm{d}t$$

Evaluate the indefinite integral first.

$$\int 27t^{3}+135t^{2}+225t+125\mathrm{d}t$$

Integrate the sum term by term.

$$\int 27t^{3}\mathrm{d}t+\int 135t^{2}\mathrm{d}t+\int 225t\mathrm{d}t+\int 125\mathrm{d}t$$

Factor out the constant in each of the terms.

$$27\int t^{3}\mathrm{d}t+135\int t^{2}\mathrm{d}t+225\int t\mathrm{d}t+\int 125\mathrm{d}t$$

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}.

$$\frac{27t^{4}}{4}+135\int t^{2}\mathrm{d}t+225\int t\mathrm{d}t+\int 125\mathrm{d}t$$

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}.

$$\frac{27t^{4}}{4}+45t^{3}+225\int t\mathrm{d}t+\int 125\mathrm{d}t$$

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}.

$$\frac{27t^{4}}{4}+45t^{3}+\frac{225t^{2}}{2}+\int 125\mathrm{d}t$$

Find the integral of 125 using the table of common integrals rule \int a\mathrm{d}t=at.

$$\frac{27t^{4}}{4}+45t^{3}+\frac{225t^{2}}{2}+125t$$

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.

$$\frac{27}{4}\times 2^{4}+45\times 2^{3}+\frac{225}{2}\times 2^{2}+125\times 2-\left(\frac{27}{4}\times 1^{4}+45\times 1^{3}+\frac{225}{2}\times 1^{2}+125\times 1\right)$$

Simplify.

$$\frac{3515}{4}$$