Solve ∫ (from 0 to 3) of (9.1x+19.8)(-6-x) wrt x

$$\int_{ 0 }^{ 3 } (9.1x+19.8)(-6-x) d x$$

Evaluate

$$-773.1$$

Solution Steps

Apply the distributive property by multiplying each term of 9.1x+19.8 by each term of -6-x.

$$\int _{0}^{3}-54.6x-9.1x^{2}-118.8-19.8x\mathrm{d}x$$

Combine -54.6x and -19.8x to get -74.4x.

$$\int _{0}^{3}-74.4x-9.1x^{2}-118.8\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -\frac{372x}{5}-\frac{91x^{2}}{10}-118.8\mathrm{d}x$$

Integrate the sum term by term.

$$\int -\frac{372x}{5}\mathrm{d}x+\int -\frac{91x^{2}}{10}\mathrm{d}x+\int -118.8\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\frac{372\int x\mathrm{d}x}{5}-\frac{91\int x^{2}\mathrm{d}x}{10}+\int -118.8\mathrm{d}x$$

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 -74.4 times \frac{x^{2}}{2}.

$$-\frac{186x^{2}}{5}-\frac{91\int x^{2}\mathrm{d}x}{10}+\int -118.8\mathrm{d}x$$

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 -9.1 times \frac{x^{3}}{3}.

$$-\frac{186x^{2}}{5}-\frac{91x^{3}}{30}+\int -118.8\mathrm{d}x$$

Find the integral of -118.8 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{186x^{2}}{5}-\frac{91x^{3}}{30}-\frac{594x}{5}$$

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.

$$-\frac{186}{5}\times 3^{2}-\frac{91}{30}\times 3^{3}-118.8\times 3-\left(-\frac{186}{5}\times 0^{2}-\frac{91}{30}\times 0^{3}-118.8\times 0\right)$$

Simplify.

$$-\frac{7731}{10}$$