Solve ∫ (from -1 to 1) of x^3+6x^2+11x+6 wrt x

$$\int_{ -1 }^{ 1 } { x }^{ 3 } +6 { x }^{ 2 } +11x+6 d x$$

Evaluate

$$16$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{3}+6x^{2}+11x+6\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int 6x^{2}\mathrm{d}x+\int 11x\mathrm{d}x+\int 6\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x+6\int x^{2}\mathrm{d}x+11\int x\mathrm{d}x+\int 6\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}+6\int x^{2}\mathrm{d}x+11\int x\mathrm{d}x+\int 6\mathrm{d}x$$

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 6 times \frac{x^{3}}{3}.

$$\frac{x^{4}}{4}+2x^{3}+11\int x\mathrm{d}x+\int 6\mathrm{d}x$$

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 11 times \frac{x^{2}}{2}.

$$\frac{x^{4}}{4}+2x^{3}+\frac{11x^{2}}{2}+\int 6\mathrm{d}x$$

Find the integral of 6 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}+2x^{3}+\frac{11x^{2}}{2}+6x$$

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.

$$\frac{1^{4}}{4}+2\times 1^{3}+\frac{11}{2}\times 1^{2}+6\times 1-\left(\frac{\left(-1\right)^{4}}{4}+2\left(-1\right)^{3}+\frac{11}{2}\left(-1\right)^{2}+6\left(-1\right)\right)$$

Simplify.

$$16$$