Solve ∫ (from - 1 to 1) of x^3+6x^2+1

$$\int _ { - 1 } ^ { 1 } x ^ { 3 } + 6 x ^ { 2 } + 1$$

Evaluate

$$6$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{3}+6x^{2}+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int 6x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x+6\int x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}+6\int x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 6 times \frac{x^{3}}{3}.

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

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

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

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.

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

Simplify.

$$6$$