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

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

Evaluate

$$-\frac{5}{6}\approx -0.833333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

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

Find the integral of -4 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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}.

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

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

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

Simplify.

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

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.

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

Simplify.

$$-\frac{5}{6}$$