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

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

Evaluate

$$\frac{40}{3}\approx 13.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

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

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

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

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.

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

Simplify.

$$\frac{40}{3}$$