Solve ∫ (from 0 to 2) of (-x(7-x))(-0.1*x) wrt x

$$\int_{ 0 }^{ 2 } (-x(7-x))(-0.1 \times x) d x$$

Evaluate

$$\frac{22}{15}\approx 1.466666667$$

Solution Steps

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

$$\int _{0}^{2}\left(7\left(-x\right)-\left(-x\right)x\right)\left(-0.1\right)x\mathrm{d}x$$

Multiply -1 and -1 to get 1.

$$\int _{0}^{2}\left(7\left(-x\right)+xx\right)\left(-0.1\right)x\mathrm{d}x$$

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

$$\int _{0}^{2}\left(7\left(-x\right)+x^{2}\right)\left(-0.1\right)x\mathrm{d}x$$

Use the distributive property to multiply 7\left(-x\right)+x^{2} by -0.1.

$$\int _{0}^{2}\left(-0.7\left(-x\right)-0.1x^{2}\right)x\mathrm{d}x$$

Multiply -0.7 and -1 to get 0.7.

$$\int _{0}^{2}\left(0.7x-0.1x^{2}\right)x\mathrm{d}x$$

Use the distributive property to multiply 0.7x-0.1x^{2} by x.

$$\int _{0}^{2}0.7x^{2}-0.1x^{3}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{7x^{2}-x^{3}}{10}\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{7x^{2}}{10}\mathrm{d}x+\int -\frac{x^{3}}{10}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{7\int x^{2}\mathrm{d}x-\int x^{3}\mathrm{d}x}{10}$$

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

$$\frac{7x^{3}}{30}-\frac{\int x^{3}\mathrm{d}x}{10}$$

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 -0.1 times \frac{x^{4}}{4}.

$$\frac{7x^{3}}{30}-\frac{x^{4}}{40}$$

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.

$$\frac{7}{30}\times 2^{3}-\frac{2^{4}}{40}-\left(\frac{7}{30}\times 0^{3}-\frac{0^{4}}{40}\right)$$

Simplify.

$$\frac{22}{15}$$