Solve ∫ (from - 1 to 2) of (2x+4)-(2x^2)] wrt x

$$\int _ { - 1 } ^ { 2 } ( 2 x + 4 ) - ( 2 x ^ { 2 } ) ] d x$$

Evaluate

$$9$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

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

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}.

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

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.

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

Simplify.

$$9$$