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

$$\int_{ 1 }^{ 3 } { x }^{ 2 } +4x-2 d x =$$

Evaluate

$$\frac{62}{3}\approx 20.666666667$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

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

Find the integral of -2 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$\frac{62}{3}$$