Solve ∫ (from 0 to 4) of (4.4x-4)(-0.2-0.1x) wrt x

$$\int_{ 0 }^{ 4 } (4.4x-4)(-0.2-0.1x) d x$$

Evaluate

$$-\frac{752}{75}\approx -10.026666667$$

Solution Steps

Apply the distributive property by multiplying each term of 4.4x-4 by each term of -0.2-0.1x.

$$\int _{0}^{4}-0.88x-0.44x^{2}+0.8+0.4x\mathrm{d}x$$

Combine -0.88x and 0.4x to get -0.48x.

$$\int _{0}^{4}-0.48x-0.44x^{2}+0.8\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -\frac{12x}{25}-\frac{11x^{2}}{25}+0.8\mathrm{d}x$$

Integrate the sum term by term.

$$\int -\frac{12x}{25}\mathrm{d}x+\int -\frac{11x^{2}}{25}\mathrm{d}x+\int 0.8\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\frac{12\int x\mathrm{d}x}{25}-\frac{11\int x^{2}\mathrm{d}x}{25}+\int 0.8\mathrm{d}x$$

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 -0.48 times \frac{x^{2}}{2}.

$$-\frac{6x^{2}}{25}-\frac{11\int x^{2}\mathrm{d}x}{25}+\int 0.8\mathrm{d}x$$

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 -0.44 times \frac{x^{3}}{3}.

$$-\frac{6x^{2}}{25}-\frac{11x^{3}}{75}+\int 0.8\mathrm{d}x$$

Find the integral of 0.8 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{6x^{2}}{25}-\frac{11x^{3}}{75}+\frac{4x}{5}$$

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.

$$-\frac{6}{25}\times 4^{2}-\frac{11}{75}\times 4^{3}+0.8\times 4-\left(-\frac{6}{25}\times 0^{2}-\frac{11}{75}\times 0^{3}+0.8\times 0\right)$$

Simplify.

$$-\frac{752}{75}$$