Solve ∫ (from 0 to 4) of 3(6t-3)(3t^2-8t-3)dt

$$\int _ { 0 } ^ { 4 } 3 ( 6 t - 3 ) ( 3 t ^ { 2 } - 8 t - 3 ) d t$$

Evaluate

$$60$$

Solution Steps

Use the distributive property to multiply 3 by 6t-3.

$$\int _{0}^{4}\left(18t-9\right)\left(3t^{2}-8t-3\right)\mathrm{d}t$$

Use the distributive property to multiply 18t-9 by 3t^{2}-8t-3 and combine like terms.

$$\int _{0}^{4}54t^{3}-171t^{2}+18t+27\mathrm{d}t$$

Evaluate the indefinite integral first.

$$\int 54t^{3}-171t^{2}+18t+27\mathrm{d}t$$

Integrate the sum term by term.

$$\int 54t^{3}\mathrm{d}t+\int -171t^{2}\mathrm{d}t+\int 18t\mathrm{d}t+\int 27\mathrm{d}t$$

Factor out the constant in each of the terms.

$$54\int t^{3}\mathrm{d}t-171\int t^{2}\mathrm{d}t+18\int t\mathrm{d}t+\int 27\mathrm{d}t$$

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

$$\frac{27t^{4}}{2}-171\int t^{2}\mathrm{d}t+18\int t\mathrm{d}t+\int 27\mathrm{d}t$$

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 -171 times \frac{t^{3}}{3}.

$$\frac{27t^{4}}{2}-57t^{3}+18\int t\mathrm{d}t+\int 27\mathrm{d}t$$

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

$$\frac{27t^{4}}{2}-57t^{3}+9t^{2}+\int 27\mathrm{d}t$$

Find the integral of 27 using the table of common integrals rule \int a\mathrm{d}t=at.

$$\frac{27t^{4}}{2}-57t^{3}+9t^{2}+27t$$

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.

$$\frac{27}{2}\times 4^{4}-57\times 4^{3}+27\times 4+9\times 4^{2}-\left(\frac{27}{2}\times 0^{4}-57\times 0^{3}+27\times 0+9\times 0^{2}\right)$$

Simplify.

$$60$$