Solve ∫ (from 0 to 2) of (-3.6x+0.5x^2)(-0.1x) wrt x

$$\int_{ 0 }^{ 2 } (-3.6x+0.5 { x }^{ 2 } )(-0.1x) d x$$

Evaluate

$$0.76$$

Solution Steps

Use the distributive property to multiply -3.6x+0.5x^{2} by -0.1.

$$\int _{0}^{2}\left(0.36x-0.05x^{2}\right)x\mathrm{d}x$$

Use the distributive property to multiply 0.36x-0.05x^{2} by x.

$$\int _{0}^{2}0.36x^{2}-0.05x^{3}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{9x^{2}}{25}-\frac{x^{3}}{20}\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{9x^{2}}{25}\mathrm{d}x+\int -\frac{x^{3}}{20}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{9\int x^{2}\mathrm{d}x}{25}-\frac{\int x^{3}\mathrm{d}x}{20}$$

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 0.36 times \frac{x^{3}}{3}.

$$\frac{3x^{3}}{25}-\frac{\int x^{3}\mathrm{d}x}{20}$$

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 -0.05 times \frac{x^{4}}{4}.

$$\frac{3x^{3}}{25}-\frac{x^{4}}{80}$$

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.

$$\frac{3}{25}\times 2^{3}-\frac{2^{4}}{80}-\left(\frac{3}{25}\times 0^{3}-\frac{0^{4}}{80}\right)$$

Simplify.

$$\frac{19}{25}$$