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

$$\int _ { - 1 } ^ { 2 } ( x ^ { 2 } - 4 x + 3 ) d x =$$

Evaluate

$$6$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}.

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

Find the integral of 3 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$6$$