Solve ∫ (from - 4 to 4) of (-2x^2+3x) wrt x

$$\int _ { - 4 } ^ { 4 } ( - 2 x ^ { 2 } + 3 x ) d x$$

Evaluate

$$-\frac{256}{3}\approx -85.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}. Multiply 3 times \frac{x^{2}}{2}.

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

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.

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

Simplify.

$$-\frac{256}{3}$$