Solve ∫ (from 0 to 4) of (12.8-0.6x)(0.8-0.1x) wrt x

$$\int_{ 0 }^{ 4 } (12.8-0.6x)(0.8-0.1x) d x$$

Evaluate

$$28.16$$

Solution Steps

Apply the distributive property by multiplying each term of 12.8-0.6x by each term of 0.8-0.1x.

$$\int _{0}^{4}10.24-1.28x-0.48x+0.06x^{2}\mathrm{d}x$$

Combine -1.28x and -0.48x to get -1.76x.

$$\int _{0}^{4}10.24-1.76x+0.06x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 10.24-\frac{44x}{25}+\frac{3x^{2}}{50}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 10.24\mathrm{d}x+\int -\frac{44x}{25}\mathrm{d}x+\int \frac{3x^{2}}{50}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 10.24\mathrm{d}x-\frac{44\int x\mathrm{d}x}{25}+\frac{3\int x^{2}\mathrm{d}x}{50}$$

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

$$\frac{256x}{25}-\frac{44\int x\mathrm{d}x}{25}+\frac{3\int x^{2}\mathrm{d}x}{50}$$

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 -1.76 times \frac{x^{2}}{2}.

$$\frac{256x}{25}-\frac{22x^{2}}{25}+\frac{3\int x^{2}\mathrm{d}x}{50}$$

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

$$\frac{256x}{25}-\frac{22x^{2}}{25}+\frac{x^{3}}{50}$$

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.

$$10.24\times 4-\frac{22}{25}\times 4^{2}+\frac{4^{3}}{50}-\left(10.24\times 0-\frac{22}{25}\times 0^{2}+\frac{0^{3}}{50}\right)$$

Simplify.

$$\frac{704}{25}$$