Solve ∫ (from 0 to 4) of (-10.4-0.6x)(0.4x-2.4) wrt x

$$\int_{ 0 }^{ 4 } (-10.4-0.6x)(0.4x-2.4) d x$$

Evaluate

$$72.96$$

Solution Steps

Apply the distributive property by multiplying each term of -10.4-0.6x by each term of 0.4x-2.4.

$$\int _{0}^{4}-4.16x+24.96-0.24x^{2}+1.44x\mathrm{d}x$$

Combine -4.16x and 1.44x to get -2.72x.

$$\int _{0}^{4}-2.72x+24.96-0.24x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{-68x+624-6x^{2}}{25}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -\frac{68x}{25}\mathrm{d}x+\int 24.96\mathrm{d}x+\int -\frac{6x^{2}}{25}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\frac{68\int x\mathrm{d}x}{25}+\int 24.96\mathrm{d}x-\frac{6\int x^{2}\mathrm{d}x}{25}$$

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 -2.72 times \frac{x^{2}}{2}.

$$-\frac{34x^{2}}{25}+\int 24.96\mathrm{d}x-\frac{6\int x^{2}\mathrm{d}x}{25}$$

Find the integral of 24.96 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{-34x^{2}+624x-6\int x^{2}\mathrm{d}x}{25}$$

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}.

$$\frac{-34x^{2}+624x-2x^{3}}{25}$$

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.

$$-\frac{34}{25}\times 4^{2}+24.96\times 4-\frac{2}{25}\times 4^{3}-\left(-\frac{34}{25}\times 0^{2}+24.96\times 0-\frac{2}{25}\times 0^{3}\right)$$

Simplify.

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