Solve ∫ (from 0 to 2) of (2.4+2.4x+0.6x^2)(x) wrt x

$$\int_{ 0 }^{ 2 } (2.4+2.4x+0.6 { x }^{ 2 } )(x) d x$$

Evaluate

$$13.6$$

Solution Steps

Use the distributive property to multiply 2.4+2.4x+0.6x^{2} by x.

$$\int _{0}^{2}2.4x+2.4x^{2}+0.6x^{3}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{12x+12x^{2}+3x^{3}}{5}\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{12x}{5}\mathrm{d}x+\int \frac{12x^{2}}{5}\mathrm{d}x+\int \frac{3x^{3}}{5}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{12\int x\mathrm{d}x+12\int x^{2}\mathrm{d}x+3\int x^{3}\mathrm{d}x}{5}$$

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 2.4 times \frac{x^{2}}{2}.

$$\frac{6x^{2}+12\int x^{2}\mathrm{d}x+3\int x^{3}\mathrm{d}x}{5}$$

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

$$\frac{6x^{2}+4x^{3}+3\int x^{3}\mathrm{d}x}{5}$$

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 0.6 times \frac{x^{4}}{4}.

$$\frac{6x^{2}}{5}+\frac{4x^{3}}{5}+\frac{3x^{4}}{20}$$

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.

$$\frac{6}{5}\times 2^{2}+\frac{4}{5}\times 2^{3}+\frac{3}{20}\times 2^{4}-\left(\frac{6}{5}\times 0^{2}+\frac{4}{5}\times 0^{3}+\frac{3}{20}\times 0^{4}\right)$$

Simplify.

$$\frac{68}{5}$$