Solve ∫ (from 0 to 0.5) of 12t^2-30t+12dt

$$\int_{ 0 }^{ 0.5 } 12 { t }^{ 2 } -30t+12 d t$$

Evaluate

$$2.75$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 12t^{2}-30t+12\mathrm{d}t$$

Integrate the sum term by term.

$$\int 12t^{2}\mathrm{d}t+\int -30t\mathrm{d}t+\int 12\mathrm{d}t$$

Factor out the constant in each of the terms.

$$12\int t^{2}\mathrm{d}t-30\int t\mathrm{d}t+\int 12\mathrm{d}t$$

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

$$4t^{3}-30\int t\mathrm{d}t+\int 12\mathrm{d}t$$

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 -30 times \frac{t^{2}}{2}.

$$4t^{3}-15t^{2}+\int 12\mathrm{d}t$$

Find the integral of 12 using the table of common integrals rule \int a\mathrm{d}t=at.

$$4t^{3}-15t^{2}+12t$$

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.

$$4\times 0.5^{3}-15\times 0.5^{2}+12\times 0.5-\left(4\times 0^{3}-15\times 0^{2}+12\times 0\right)$$

Simplify.

$$2.75$$