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

$$\int _ { - \pi } ^ { 2 } [ - 4 x + 4 + x ^ { 2 } ) d x$$

Evaluate

$$\frac{\pi ^{3}}{3}+2\pi ^{2}+4\pi +\frac{8}{3}\approx 45.307671643$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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 -4 times \frac{x^{2}}{2}.

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

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

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

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}.

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

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.

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

Simplify.

$$\frac{8}{3}+2\pi ^{2}+4\pi +\frac{\pi ^{3}}{3}$$