Solve ∫ (from 0 to 4) of (-10.4-0.6x)(0.4+0.1x) wrt x

$$\int_{ 0 }^{ 4 } (-10.4-0.6x)(0.4+0.1x) d x$$

Evaluate

$$-28.16$$

Solution Steps

Apply the distributive property by multiplying each term of -10.4-0.6x by each term of 0.4+0.1x.

$$\int _{0}^{4}-4.16-1.04x-0.24x-0.06x^{2}\mathrm{d}x$$

Combine -1.04x and -0.24x to get -1.28x.

$$\int _{0}^{4}-4.16-1.28x-0.06x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -4.16-\frac{32x}{25}-\frac{3x^{2}}{50}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -4.16\mathrm{d}x+\int -\frac{32x}{25}\mathrm{d}x+\int -\frac{3x^{2}}{50}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int -4.16\mathrm{d}x-\frac{32\int x\mathrm{d}x}{25}-\frac{3\int x^{2}\mathrm{d}x}{50}$$

Find the integral of -4.16 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{104x}{25}-\frac{32\int x\mathrm{d}x}{25}-\frac{3\int x^{2}\mathrm{d}x}{50}$$

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 -1.28 times \frac{x^{2}}{2}.

$$-\frac{104x}{25}-\frac{16x^{2}}{25}-\frac{3\int x^{2}\mathrm{d}x}{50}$$

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}.

$$-\frac{104x}{25}-\frac{16x^{2}}{25}-\frac{x^{3}}{50}$$

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.

$$-4.16\times 4-\frac{16}{25}\times 4^{2}-\frac{4^{3}}{50}-\left(-4.16\times 0-\frac{16}{25}\times 0^{2}-\frac{0^{3}}{50}\right)$$

Simplify.

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