Solve ∫ (from 0 to 4) of (-0.1x)(x^2-8x) wrt x

$$\int_{ 0 }^{ 4 } (-0.1x)( { x }^{ 2 } -8x) d x$$

Evaluate

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

Solution Steps

Use the distributive property to multiply -0.1x by x^{2}-8x.

$$\int _{0}^{4}-0.1x^{3}+0.8x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -\frac{x^{3}}{10}+\frac{4x^{2}}{5}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -\frac{x^{3}}{10}\mathrm{d}x+\int \frac{4x^{2}}{5}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\frac{\int x^{3}\mathrm{d}x}{10}+\frac{4\int x^{2}\mathrm{d}x}{5}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -0.1 times \frac{x^{4}}{4}.

$$-\frac{x^{4}}{40}+\frac{4\int x^{2}\mathrm{d}x}{5}$$

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}. Multiply 0.8 times \frac{x^{3}}{3}.

$$-\frac{x^{4}}{40}+\frac{4x^{3}}{15}$$

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.

$$-\frac{4^{4}}{40}+\frac{4}{15}\times 4^{3}-\left(-\frac{0^{4}}{40}+\frac{4}{15}\times 0^{3}\right)$$

Simplify.

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