Solve ∫ (from - 1 to 1) of (x^3+x+1)

$$\int _ { - 1 } ^ { 1 } ( x ^ { 3 } + x + 1 )$$

Evaluate

$$2$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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}.

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

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}.

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

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

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

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.

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

Simplify.

$$2$$