Solve ∫ (from - 1 to 1) of 3x+x^2

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

Evaluate

$$\frac{2}{3}\approx 0.666666667$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}.

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

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.

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

Simplify.

$$\frac{2}{3}$$