Solve ∫ (from 0 to 1) of (x^2+2-(-x))

$$\int _ { 0 } ^ { 1 } ( x ^ { 2 } + 2 - ( - x ) )$$

Evaluate

$$\frac{17}{6}\approx 2.833333333$$

Solution Steps

Multiply -1 and -1 to get 1.

$$\int _{0}^{1}x^{2}+2+x\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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}.

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

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

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

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}.

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

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.

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

Simplify.

$$\frac{17}{6}$$