Solve ∫ (from 0 to 4) of (3x+1x^2)(-0.125(x)) wrt x

$$\int_{ 0 }^{ 4 } (3x+1 { x }^{ 2 } )(-0.125(x)) d x$$

Evaluate

$$-16$$

Solution Steps

Use the distributive property to multiply 3x+1x^{2} by -0.125.

$$\int _{0}^{4}\left(-0.375x-0.125x^{2}\right)x\mathrm{d}x$$

Use the distributive property to multiply -0.375x-0.125x^{2} by x.

$$\int _{0}^{4}-0.375x^{2}-0.125x^{3}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{-3x^{2}-x^{3}}{8}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -\frac{3x^{2}}{8}\mathrm{d}x+\int -\frac{x^{3}}{8}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{-3\int x^{2}\mathrm{d}x-\int x^{3}\mathrm{d}x}{8}$$

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 -0.375 times \frac{x^{3}}{3}.

$$\frac{-x^{3}-\int x^{3}\mathrm{d}x}{8}$$

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 -0.125 times \frac{x^{4}}{4}.

$$-\frac{x^{3}}{8}-\frac{x^{4}}{32}$$

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.

$$-\frac{4^{3}}{8}-\frac{4^{4}}{32}-\left(-\frac{0^{3}}{8}-\frac{0^{4}}{32}\right)$$

Simplify.

$$-16$$