Solve ∫ (from 0 to 2) of 2x^2+x

$$\int _ { 0 } ^ { 2 } 2 x ^ { 2 } + x$$

Evaluate

$$\frac{22}{3}\approx 7.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 2 times \frac{x^{3}}{3}.

$$\frac{2x^{3}}{3}+\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}.

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

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.

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

Simplify.

$$\frac{22}{3}$$