Solve ∫ (from 0 to 8) of (-16-8x+80/3+x)(1) wrt x

$$\int_{ 0 }^{ 8 } (-16-8x+ \frac{ 80 }{ 3 } +x)(1) d x$$

Evaluate

$$-\frac{416}{3}\approx -138.666666667$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -16-8x+\frac{80}{3}+x\mathrm{d}x$$

Integrate the sum term by term.

$$\int -16\mathrm{d}x+\int -8x\mathrm{d}x+\int \frac{80}{3}\mathrm{d}x+\int x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int -16\mathrm{d}x-8\int x\mathrm{d}x+\int \frac{80}{3}\mathrm{d}x+\int x\mathrm{d}x$$

Find the integral of -16 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-16x-8\int x\mathrm{d}x+\int \frac{80}{3}\mathrm{d}x+\int x\mathrm{d}x$$

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 -8 times \frac{x^{2}}{2}.

$$-16x-4x^{2}+\int \frac{80}{3}\mathrm{d}x+\int x\mathrm{d}x$$

Find the integral of \frac{80}{3} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-16x-4x^{2}+\frac{80x}{3}+\int x\mathrm{d}x$$

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}.

$$-16x-4x^{2}+\frac{80x}{3}+\frac{x^{2}}{2}$$

Simplify.

$$\frac{32x}{3}-\frac{7x^{2}}{2}$$

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.

$$\frac{32}{3}\times 8-\frac{7}{2}\times 8^{2}-\left(\frac{32}{3}\times 0-\frac{7}{2}\times 0^{2}\right)$$

Simplify.

$$-\frac{416}{3}$$