Solve ∫ (from 4 to 8) of (x^2-12x+32) wrt x

$$\int _ { 4 } ^ { 8 } ( x ^ { 2 } - 12 x + 32 ) d x$$

Evaluate

$$-\frac{32}{3}\approx -10.666666667$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}-12x+32\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int -12x\mathrm{d}x+\int 32\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x-12\int x\mathrm{d}x+\int 32\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}-12\int x\mathrm{d}x+\int 32\mathrm{d}x$$

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 -12 times \frac{x^{2}}{2}.

$$\frac{x^{3}}{3}-6x^{2}+\int 32\mathrm{d}x$$

Find the integral of 32 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}-6x^{2}+32x$$

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.

$$\frac{8^{3}}{3}-6\times 8^{2}+32\times 8-\left(\frac{4^{3}}{3}-6\times 4^{2}+32\times 4\right)$$

Simplify.

$$-\frac{32}{3}$$