Solve ∫ (from 0 to 4) of (-3.625x+1.5x^2)(-0.5x) wrt x

$$\int_{ 0 }^{ 4 } (-3.625x+1.5 { x }^{ 2 } )(-0.5x) d x$$

Evaluate

$$-\frac{28}{3}\approx -9.333333333$$

Solution Steps

Use the distributive property to multiply -3.625x+1.5x^{2} by -0.5.

$$\int _{0}^{4}\left(1.8125x-0.75x^{2}\right)x\mathrm{d}x$$

Use the distributive property to multiply 1.8125x-0.75x^{2} by x.

$$\int _{0}^{4}1.8125x^{2}-0.75x^{3}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{29x^{2}}{16}-\frac{3x^{3}}{4}\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{29x^{2}}{16}\mathrm{d}x+\int -\frac{3x^{3}}{4}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{29\int x^{2}\mathrm{d}x}{16}-\frac{3\int x^{3}\mathrm{d}x}{4}$$

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 1.8125 times \frac{x^{3}}{3}.

$$\frac{29x^{3}}{48}-\frac{3\int x^{3}\mathrm{d}x}{4}$$

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 -0.75 times \frac{x^{4}}{4}.

$$\frac{29x^{3}}{48}-\frac{3x^{4}}{16}$$

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.

$$\frac{29}{48}\times 4^{3}-\frac{3}{16}\times 4^{4}-\left(\frac{29}{48}\times 0^{3}-\frac{3}{16}\times 0^{4}\right)$$

Simplify.

$$-\frac{28}{3}$$