Solve ∫ (from 0 to 4) of (-x(7-x))(-0.6x) wrt x

$$\int_{ 0 }^{ 4 } (-x(7-x))(-0.6x) d x$$

Evaluate

$$51.2$$

Solution Steps

Use the distributive property to multiply -x by 7-x.

$$\int _{0}^{4}\left(7\left(-x\right)-\left(-x\right)x\right)\left(-0.6\right)x\mathrm{d}x$$

Multiply -1 and -1 to get 1.

$$\int _{0}^{4}\left(7\left(-x\right)+xx\right)\left(-0.6\right)x\mathrm{d}x$$

Multiply x and x to get x^{2}.

$$\int _{0}^{4}\left(7\left(-x\right)+x^{2}\right)\left(-0.6\right)x\mathrm{d}x$$

Use the distributive property to multiply 7\left(-x\right)+x^{2} by -0.6.

$$\int _{0}^{4}\left(-4.2\left(-x\right)-0.6x^{2}\right)x\mathrm{d}x$$

Multiply -4.2 and -1 to get 4.2.

$$\int _{0}^{4}\left(4.2x-0.6x^{2}\right)x\mathrm{d}x$$

Use the distributive property to multiply 4.2x-0.6x^{2} by x.

$$\int _{0}^{4}4.2x^{2}-0.6x^{3}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{21x^{2}-3x^{3}}{5}\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{21x^{2}}{5}\mathrm{d}x+\int -\frac{3x^{3}}{5}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{21\int x^{2}\mathrm{d}x-3\int x^{3}\mathrm{d}x}{5}$$

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

$$\frac{7x^{3}-3\int x^{3}\mathrm{d}x}{5}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -0.6 times \frac{x^{4}}{4}.

$$\frac{7x^{3}}{5}-\frac{3x^{4}}{20}$$

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.

$$\frac{7}{5}\times 4^{3}-\frac{3}{20}\times 4^{4}-\left(\frac{7}{5}\times 0^{3}-\frac{3}{20}\times 0^{4}\right)$$

Simplify.

$$\frac{256}{5}$$