Solve ∫ (from - 1 to 4) of x^2-6x+4 wrt x

$$\int _ { - 1 } ^ { 4 } x ^ { 2 } - 6 x + 4 d x$$

Evaluate

$$-\frac{10}{3}\approx -3.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}-6x+4\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int -6x\mathrm{d}x+\int 4\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x+\int 4\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}-6\int x\mathrm{d}x+\int 4\mathrm{d}x$$

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 -6 times \frac{x^{2}}{2}.

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

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

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

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.

$$\frac{4^{3}}{3}-3\times 4^{2}+4\times 4-\left(\frac{\left(-1\right)^{3}}{3}-3\left(-1\right)^{2}+4\left(-1\right)\right)$$

Simplify.

$$-\frac{10}{3}$$