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

$$\int_{ -2 }^{ 4 } (- \frac{ 1 }{ 2 } { x }^{ 2 } +x+4) d x$$

Evaluate

$$18$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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 -\frac{1}{2} times \frac{x^{3}}{3}.

$$-\frac{x^{3}}{6}+\int x\mathrm{d}x+\int 4\mathrm{d}x$$

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}.

$$-\frac{x^{3}}{6}+\frac{x^{2}}{2}+\int 4\mathrm{d}x$$

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

$$-\frac{x^{3}}{6}+\frac{x^{2}}{2}+4x$$

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.

$$-\frac{4^{3}}{6}+\frac{4^{2}}{2}+4\times 4-\left(-\frac{\left(-2\right)^{3}}{6}+\frac{\left(-2\right)^{2}}{2}+4\left(-2\right)\right)$$

Simplify.

$$18$$