Solve ∫ (from 0 to 4) of (0.46x-0.05x^3) wrt x

$$\int_{ 0 }^{ 4 } (0.46x-0.05 { x }^{ 3 } ) d x$$

Evaluate

$$0.48$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{23x}{50}-\frac{x^{3}}{20}\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{23x}{50}\mathrm{d}x+\int -\frac{x^{3}}{20}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{23\int x\mathrm{d}x}{50}-\frac{\int x^{3}\mathrm{d}x}{20}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 0.46 times \frac{x^{2}}{2}.

$$\frac{23x^{2}}{100}-\frac{\int x^{3}\mathrm{d}x}{20}$$

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}.

$$\frac{23x^{2}}{100}-\frac{x^{4}}{80}$$

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.

$$\frac{23}{100}\times 4^{2}-\frac{4^{4}}{80}-\left(\frac{23}{100}\times 0^{2}-\frac{0^{4}}{80}\right)$$

Simplify.

$$\frac{12}{25}$$