Solve ∫ (from 0 to 6) of 4+0.5t^2dt

$$\int _ { 0 } ^ { 6 } 4 + 0.5 t ^ { 2 } d t$$

Evaluate

$$60$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}t=at.

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

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

$$4t+\frac{t^{3}}{6}$$

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.

$$4\times 6+\frac{6^{3}}{6}-\left(4\times 0+\frac{0^{3}}{6}\right)$$

Simplify.

$$60$$