Solve ∫ (from 0 to 8) of (6.67x-x^2)(-0.083x) wrt x

$$\int_{ 0 }^{ 8 } (6.67x- { x }^{ 2 } )(-0.083x) d x$$

Evaluate

$$-\frac{88976}{9375}\approx -9.490773333$$

Solution Steps

Use the distributive property to multiply 6.67x-x^{2} by -0.083.

$$\int _{0}^{8}\left(-0.55361x+0.083x^{2}\right)x\mathrm{d}x$$

Use the distributive property to multiply -0.55361x+0.083x^{2} by x.

$$\int _{0}^{8}-0.55361x^{2}+0.083x^{3}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -\frac{55361x^{2}}{100000}+\frac{83x^{3}}{1000}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -\frac{55361x^{2}}{100000}\mathrm{d}x+\int \frac{83x^{3}}{1000}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\frac{55361\int x^{2}\mathrm{d}x}{100000}+\frac{83\int x^{3}\mathrm{d}x}{1000}$$

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 -0.55361 times \frac{x^{3}}{3}.

$$-\frac{55361x^{3}}{300000}+\frac{83\int x^{3}\mathrm{d}x}{1000}$$

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 0.083 times \frac{x^{4}}{4}.

$$-\frac{55361x^{3}}{300000}+\frac{83x^{4}}{4000}$$

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.

$$-\frac{55361}{300000}\times 8^{3}+\frac{83}{4000}\times 8^{4}-\left(-\frac{55361}{300000}\times 0^{3}+\frac{83}{4000}\times 0^{4}\right)$$

Simplify.

$$-\frac{88976}{9375}$$