Solve ∫ (from 0 to 4) of (3x+1x^2)(-0.5x) wrt x

$$\int_{ 0 }^{ 4 } (3x+1 { x }^{ 2 } )(-0.5x) d x$$

Evaluate

$$-64$$

Solution Steps

Use the distributive property to multiply 3x+1x^{2} by -0.5.

$$\int _{0}^{4}\left(-1.5x-0.5x^{2}\right)x\mathrm{d}x$$

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

$$\int _{0}^{4}-1.5x^{2}-0.5x^{3}\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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 -1.5 times \frac{x^{3}}{3}.

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

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 -0.5 times \frac{x^{4}}{4}.

$$-\frac{x^{3}}{2}-\frac{x^{4}}{8}$$

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.

$$-\frac{4^{3}}{2}-\frac{4^{4}}{8}-\left(-\frac{0^{3}}{2}-\frac{0^{4}}{8}\right)$$

Simplify.

$$-64$$