Solve ∫ (from - 8 to 8) of x^4+x^3+2x+1 wrt x

$$\int _ { - 8 } ^ { 8 } x ^ { 4 } + x ^ { 3 } + 2 x + 1 d x$$

Evaluate

$$\frac{65616}{5}=13123.2$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{4}+x^{3}+2x+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{4}\mathrm{d}x+\int x^{3}\mathrm{d}x+\int 2x\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{4}\mathrm{d}x+\int x^{3}\mathrm{d}x+2\int x\mathrm{d}x+\int 1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}.

$$\frac{x^{5}}{5}+\int x^{3}\mathrm{d}x+2\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}+\frac{x^{4}}{4}+2\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}+\frac{x^{4}}{4}+x^{2}+\int 1\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{5}}{5}+\frac{x^{4}}{4}+x^{2}+x$$

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.

$$\frac{8^{5}}{5}+\frac{8^{4}}{4}+8^{2}+8-\left(\frac{\left(-8\right)^{5}}{5}+\frac{\left(-8\right)^{4}}{4}+\left(-8\right)^{2}-8\right)$$

Simplify.

$$\frac{65616}{5}$$