Solve ∫ (from 0 to 1) of 3t(6t^2+3)dt

$$\int _ { 0 } ^ { 1 } 3 t ( 6 t ^ { 2 } + 3 ) d t$$

Evaluate

$$9$$

Solution Steps

Use the distributive property to multiply 3t by 6t^{2}+3.

$$\int _{0}^{1}18t^{3}+9t\mathrm{d}t$$

Evaluate the indefinite integral first.

$$\int 18t^{3}+9t\mathrm{d}t$$

Integrate the sum term by term.

$$\int 18t^{3}\mathrm{d}t+\int 9t\mathrm{d}t$$

Factor out the constant in each of the terms.

$$18\int t^{3}\mathrm{d}t+9\int t\mathrm{d}t$$

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

$$\frac{9t^{4}}{2}+9\int t\mathrm{d}t$$

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

$$\frac{9t^{4}+9t^{2}}{2}$$

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.

$$\frac{9}{2}\times 1^{4}+\frac{9}{2}\times 1^{2}-\left(\frac{9}{2}\times 0^{4}+\frac{9}{2}\times 0^{2}\right)$$

Simplify.

$$9$$