Solve ∫ (from 4 to 8) of x^2-4x+x

$$\int _ { 4 } ^ { 8 } x ^ { 2 } - 4 x + x$$

Evaluate

$$\frac{232}{3}\approx 77.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}.

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

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}.

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

Simplify.

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

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.

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

Simplify.

$$\frac{232}{3}$$