Solve ∫ (from - 2 to 2) of x^3+x^2+8

$$\int _ { - 2 } ^ { 2 } x ^ { 3 } + x ^ { 2 } + 8$$

Evaluate

$$\frac{112}{3}\approx 37.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{3}+x^{2}+8\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x+\int 8\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

$$\frac{x^{4}}{4}+\int x^{2}\mathrm{d}x+\int 8\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}+\frac{x^{3}}{3}+\int 8\mathrm{d}x$$

Find the integral of 8 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}+\frac{x^{3}}{3}+8x$$

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.

$$\frac{2^{4}}{4}+\frac{2^{3}}{3}+8\times 2-\left(\frac{\left(-2\right)^{4}}{4}+\frac{\left(-2\right)^{3}}{3}+8\left(-2\right)\right)$$

Simplify.

$$\frac{112}{3}$$