Solve ∫ (from 0 to 4) of (0.6x+11.2)(0.4x+0.8) wrt x

$$\int_{ 0 }^{ 4 } (0.6x+11.2)(0.4x+0.8) d x$$

Evaluate

$$80.64$$

Solution Steps

Apply the distributive property by multiplying each term of 0.6x+11.2 by each term of 0.4x+0.8.

$$\int _{0}^{4}0.24x^{2}+0.48x+4.48x+8.96\mathrm{d}x$$

Combine 0.48x and 4.48x to get 4.96x.

$$\int _{0}^{4}0.24x^{2}+4.96x+8.96\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{6x^{2}+124x+224}{25}\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{6x^{2}}{25}\mathrm{d}x+\int \frac{124x}{25}\mathrm{d}x+\int 8.96\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{6\int x^{2}\mathrm{d}x}{25}+\frac{124\int x\mathrm{d}x}{25}+\int 8.96\mathrm{d}x$$

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

$$\frac{2x^{3}}{25}+\frac{124\int x\mathrm{d}x}{25}+\int 8.96\mathrm{d}x$$

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

$$\frac{2x^{3}}{25}+\frac{62x^{2}}{25}+\int 8.96\mathrm{d}x$$

Find the integral of 8.96 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2x^{3}+62x^{2}+224x}{25}$$

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.

$$\frac{2}{25}\times 4^{3}+\frac{62}{25}\times 4^{2}+8.96\times 4-\left(\frac{2}{25}\times 0^{3}+\frac{62}{25}\times 0^{2}+8.96\times 0\right)$$

Simplify.

$$\frac{2016}{25}$$