Solve ∫ (from - 1 to 2) of 6-x^2-(x^2-2x+2) wrt x

$$\int _ { - 1 } ^ { 2 } 6 - x ^ { 2 } - ( x ^ { 2 } - 2 x + 2 ) d x$$

Evaluate

$$9$$

Solution Steps

To find the opposite of x^{2}-2x+2, find the opposite of each term.

$$\int _{-1}^{2}6-x^{2}-x^{2}+2x-2\mathrm{d}x$$

Combine -x^{2} and -x^{2} to get -2x^{2}.

$$\int _{-1}^{2}6-2x^{2}+2x-2\mathrm{d}x$$

Subtract 2 from 6 to get 4.

$$\int _{-1}^{2}4-2x^{2}+2x\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 4-2x^{2}+2x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4\mathrm{d}x+\int -2x^{2}\mathrm{d}x+\int 2x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 4\mathrm{d}x-2\int x^{2}\mathrm{d}x+2\int x\mathrm{d}x$$

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

$$4x-2\int x^{2}\mathrm{d}x+2\int x\mathrm{d}x$$

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}.

$$4x-\frac{2x^{3}}{3}+2\int x\mathrm{d}x$$

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}. Multiply 2 times \frac{x^{2}}{2}.

$$4x-\frac{2x^{3}}{3}+x^{2}$$

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.

$$4\times 2-\frac{2}{3}\times 2^{3}+2^{2}-\left(4\left(-1\right)-\frac{2}{3}\left(-1\right)^{3}+\left(-1\right)^{2}\right)$$

Simplify.

$$9$$