Solve = ∫ (from - 1 to 2) of 2-x^2+x

$$= \int _ { - 1 } ^ { 2 } 2 - x ^ { 2 } + x$$

Evaluate

$$\frac{9}{2}=4.5$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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 -1 times \frac{x^{3}}{3}.

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

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}.

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

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.

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

Simplify.

$$\frac{9}{2}$$