Solve ∫ (from 0 to 1) of (2+t^2)dt

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

Evaluate

$$\frac{7}{3}\approx 2.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}t=at.

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

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}.

$$2t+\frac{t^{3}}{3}$$

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.

$$2\times 1+\frac{1^{3}}{3}-\left(2\times 0+\frac{0^{3}}{3}\right)$$

Simplify.

$$\frac{7}{3}$$