Solve ∫ (from 0 to 4) of (12.8-0.6x)(0.8+0.4x) wrt x

$$\int_{ 0 }^{ 4 } (12.8-0.6x)(0.8+0.4x) d x$$

Evaluate

$$72.96$$

Solution Steps

Apply the distributive property by multiplying each term of 12.8-0.6x by each term of 0.8+0.4x.

$$\int _{0}^{4}10.24+5.12x-0.48x-0.24x^{2}\mathrm{d}x$$

Combine 5.12x and -0.48x to get 4.64x.

$$\int _{0}^{4}10.24+4.64x-0.24x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{256+116x-6x^{2}}{25}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 10.24\mathrm{d}x+\int \frac{116x}{25}\mathrm{d}x+\int -\frac{6x^{2}}{25}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 10.24\mathrm{d}x+\frac{116\int x\mathrm{d}x}{25}-\frac{6\int x^{2}\mathrm{d}x}{25}$$

Find the integral of 10.24 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{256x+116\int x\mathrm{d}x-6\int x^{2}\mathrm{d}x}{25}$$

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

$$\frac{256x+58x^{2}-6\int x^{2}\mathrm{d}x}{25}$$

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}.

$$\frac{256x+58x^{2}-2x^{3}}{25}$$

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.

$$10.24\times 4+\frac{58}{25}\times 4^{2}-\frac{2}{25}\times 4^{3}-\left(10.24\times 0+\frac{58}{25}\times 0^{2}-\frac{2}{25}\times 0^{3}\right)$$

Simplify.

$$\frac{1824}{25}$$