Solve ∫ (from 0 to 1) of (6t^2+4t+6)(120t+40)dt

$$\int _ { 0 } ^ { 1 } ( 6 t ^ { 2 } + 4 t + 6 ) ( 120 t + 40 ) d t$$

Evaluate

$$1100$$

Solution Steps

Use the distributive property to multiply 6t^{2}+4t+6 by 120t+40 and combine like terms.

$$\int _{0}^{1}720t^{3}+720t^{2}+880t+240\mathrm{d}t$$

Evaluate the indefinite integral first.

$$\int 720t^{3}+720t^{2}+880t+240\mathrm{d}t$$

Integrate the sum term by term.

$$\int 720t^{3}\mathrm{d}t+\int 720t^{2}\mathrm{d}t+\int 880t\mathrm{d}t+\int 240\mathrm{d}t$$

Factor out the constant in each of the terms.

$$720\int t^{3}\mathrm{d}t+720\int t^{2}\mathrm{d}t+880\int t\mathrm{d}t+\int 240\mathrm{d}t$$

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

$$180t^{4}+720\int t^{2}\mathrm{d}t+880\int t\mathrm{d}t+\int 240\mathrm{d}t$$

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

$$180t^{4}+240t^{3}+880\int t\mathrm{d}t+\int 240\mathrm{d}t$$

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

$$180t^{4}+240t^{3}+440t^{2}+\int 240\mathrm{d}t$$

Find the integral of 240 using the table of common integrals rule \int a\mathrm{d}t=at.

$$180t^{4}+240t^{3}+440t^{2}+240t$$

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.

$$180\times 1^{4}+240\times 1^{3}+240\times 1+440\times 1^{2}-\left(180\times 0^{4}+240\times 0^{3}+240\times 0+440\times 0^{2}\right)$$

Simplify.

$$1100$$